Chapter 3 Trigonometric Functions (Additional Questions)

To convert degrees into radian measure, we use the formula:

Radian Measure $=$ Degree Measure $\times \frac{\pi}{180^\circ}$

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Substituting the given degree measure $240^\circ$ into the formula:

Radian Measure $=$ $240^\circ \times \frac{\pi}{180^\circ}$

Radian Measure $=$ $\frac{240\pi}{180}$ radians

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Simplifying the fraction $\frac{240}{180}$:

$\frac{240}{180} = \frac{24}{18} = \frac{4 \times 6}{3 \times 6} = \frac{4}{3}$

So,

Radian Measure $=$ $\frac{4\pi}{3}$ radians

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This corresponds to option (B).

The final answer is $\frac{4\pi}{3}$ radians.

To convert radian measure into degrees, we use the formula:

Degree Measure $=$ Radian Measure $\times \frac{180^\circ}{\pi}$

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Substituting the given radian measure $\frac{7\pi}{6}$ into the formula:

Degree Measure $=$ $\frac{7\pi}{6} \times \frac{180^\circ}{\pi}$

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Cancel out $\pi$ from the numerator and the denominator:

Degree Measure $=$ $\frac{7}{\cancel{6}_{1}} \times \cancel{180^\circ}^{30}$

Degree Measure $=$ $7 \times 30^\circ$

Degree Measure $=$ $210^\circ$

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This corresponds to option (A).

The final answer is $210^\circ$.

To find the quadrant of an angle, we can find a coterminal angle that lies between $0^\circ$ and $360^\circ$.

Coterminal angles are obtained by adding or subtracting multiples of $360^\circ$.

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Given angle is $-1230^\circ$. Since it is a negative angle, we add multiples of $360^\circ$ to get a positive coterminal angle.

Let's add $4$ times $360^\circ$:

$4 \times 360^\circ = 1440^\circ$

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Adding this to the given angle:

Coterminal angle $=$ $-1230^\circ + 1440^\circ$

Coterminal angle $=$ $210^\circ$

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Now we determine the quadrant for the angle $210^\circ$.

The quadrants are defined as follows:

First Quadrant: $0^\circ < \theta < 90^\circ$

Second Quadrant: $90^\circ < \theta < 180^\circ$

Third Quadrant: $180^\circ < \theta < 270^\circ$

Fourth Quadrant: $270^\circ < \theta < 360^\circ$

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Since $180^\circ < 210^\circ < 270^\circ$, the angle $210^\circ$ lies in the Third Quadrant.

Therefore, the angle $-1230^\circ$ lies in the Third Quadrant.

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This corresponds to option (C).

The final answer is Third Quadrant.

Given:

Number of revolutions made by the wheel in 1 minute = 480 revolutions.

To Find:

The number of radians the wheel turns in one second.

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Solution:

First, we convert the time from minutes to seconds. 1 minute = 60 seconds.

The wheel makes 480 revolutions in 60 seconds.

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Next, we find the number of revolutions made by the wheel in one second.

Number of revolutions in 1 second $=$ $\frac{\text{Total revolutions}}{\text{Total time in seconds}}$

Number of revolutions in 1 second $=$ $\frac{480}{60}$

Number of revolutions in 1 second $=$ $\frac{\cancel{480}^{8}}{\cancel{60}_{1}}$

Number of revolutions in 1 second $=$ 8 revolutions.

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We know that one complete revolution corresponds to an angle of $2\pi$ radians.

So, to find the angle in radians for 8 revolutions, we multiply the number of revolutions by the angle per revolution.

Angle turned in radians $=$ Number of revolutions $\times$ Angle per revolution

Angle turned in radians $=$ $8 \times 2\pi$ radians

Angle turned in radians $=$ $16\pi$ radians.

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Thus, the wheel turns through $16\pi$ radians in one second.

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This corresponds to option (B).

The final answer is $16\pi$ radians.

Given:

Radius of the circle, $r = 25$ cm.

Length of the arc, $s = 11$ cm.

Value of $\pi = \frac{22}{7}$.

To Find:

The degree measure of the angle subtended at the centre.

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Solution:

The formula relating the arc length ($s$), radius ($r$), and the angle subtended at the centre ($\theta$) in radians is $s = r\theta$.

We can rearrange this formula to find the angle in radians:

$\theta = \frac{s}{r}$

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Substitute the given values of $s$ and $r$ into the formula:

$\theta = \frac{11 \text{ cm}}{25 \text{ cm}}$

$\theta = \frac{11}{25}$ radians

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Now, we need to convert this angle from radians to degrees. The conversion formula is:

Degree Measure $=$ Radian Measure $\times \frac{180^\circ}{\pi}$

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Substitute the value of $\theta$ in radians and the given value of $\pi = \frac{22}{7}$ into the conversion formula:

Degree Measure $=$ $\frac{11}{25} \times \frac{180^\circ}{22/7}$

Degree Measure $=$ $\frac{11}{25} \times 180^\circ \times \frac{7}{22}$

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Simplify the expression:

Degree Measure $=$ $\frac{\cancel{11}^{1}}{25} \times 180^\circ \times \frac{7}{\cancel{22}_{2}}$

Degree Measure $=$ $\frac{1}{25} \times 180^\circ \times \frac{7}{2}$

Degree Measure $=$ $\frac{1}{25} \times \cancel{180^\circ}^{90} \times \frac{7}{\cancel{2}_{1}}$

Degree Measure $=$ $\frac{90 \times 7}{25}$ degrees

Degree Measure $=$ $\frac{630}{25}$ degrees

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Further simplification of the fraction $\frac{630}{25}$:

$\frac{630}{25} = \frac{126 \times 5}{5 \times 5} = \frac{126}{5}$

$\frac{126}{5} = 25.2$

So, the degree measure is $25.2^\circ$.

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This corresponds to option (A).

The final answer is $25.2^\circ$.

Given:

Radius of the circle, $r = 14$ cm.

Central angle, $\theta = \frac{2\pi}{7}$ radians.

To Find:

The area of the sector.

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Solution:

The formula for the area of a sector of a circle with radius $r$ and central angle $\theta$ (in radians) is given by:

Area $= \frac{1}{2} r^2 \theta$

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Substitute the given values of $r$ and $\theta$ into the formula:

Area $= \frac{1}{2} \times (14 \, \text{cm})^2 \times \left(\frac{2\pi}{7}\right)$ radians

Area $= \frac{1}{2} \times 196 \, \text{cm}^2 \times \frac{2\pi}{7}$

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Simplify the expression:

Area $= \frac{1}{\cancel{2}_{1}} \times 196 \times \frac{\cancel{2}^{1}\pi}{7} \, \text{cm}^2$

Area $= 196 \times \frac{\pi}{7} \, \text{cm}^2$

Area $= \frac{196}{7} \pi \, \text{cm}^2$

Area $= 28\pi \, \text{cm}^2$

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Since the options are numerical values, we use the standard approximation for $\pi$, which is $\frac{22}{7}$.

Area $= 28 \times \frac{22}{7} \, \text{cm}^2$

Area $= \cancel{28}^{4} \times \frac{22}{\cancel{7}_{1}} \, \text{cm}^2$

Area $= 4 \times 22 \, \text{cm}^2$

Area $= 88 \, \text{cm}^2$

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This corresponds to option (C).

The final answer is $88 \, \text{cm}^2$.

To find the value of $\sin(510^\circ)$, we can express the angle in terms of a coterminal angle or a standard angle using the periodicity of the sine function.

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The sine function has a period of $360^\circ$. Therefore, $\sin(\theta + n \times 360^\circ) = \sin(\theta)$ for any integer $n$.

We can write $510^\circ$ as $1 \times 360^\circ + 150^\circ$.

So, $\sin(510^\circ) = \sin(360^\circ + 150^\circ)$.

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Using the periodicity property:

$\sin(510^\circ) = \sin(150^\circ)$

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The angle $150^\circ$ lies in the second quadrant. We can express $150^\circ$ as $180^\circ - 30^\circ$.

Using the reduction formula $\sin(180^\circ - \theta) = \sin(\theta)$ for the second quadrant:

$\sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin(30^\circ)$

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The value of $\sin(30^\circ)$ is a standard trigonometric value:

$\sin(30^\circ) = \frac{1}{2}$

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Therefore, $\sin(510^\circ) = \frac{1}{2}$.

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This corresponds to option (A).

The final answer is $\frac{1}{2}$.

Assertion (A): The value of $\cos(\frac{17\pi}{3})$ is $\frac{1}{2}$.

Let's evaluate $\cos(\frac{17\pi}{3})$. We can write the angle $\frac{17\pi}{3}$ by separating multiples of $2\pi$.

$\frac{17\pi}{3} = \frac{18\pi - \pi}{3} = \frac{18\pi}{3} - \frac{\pi}{3} = 6\pi - \frac{\pi}{3}$.

Since $6\pi = 3 \times 2\pi$, this is in the form $2n\pi - x$ where $n=3$ and $x = \frac{\pi}{3}$.

Using the property $\cos(2n\pi - x) = \cos(-x) = \cos x$, we have:

$\cos(\frac{17\pi}{3}) = \cos(6\pi - \frac{\pi}{3})$

$\cos(\frac{17\pi}{3}) = \cos(-\frac{\pi}{3})$

$\cos(\frac{17\pi}{3}) = \cos(\frac{\pi}{3})$

The value of $\cos(\frac{\pi}{3})$ is $\frac{1}{2}$.

So, $\cos(\frac{17\pi}{3}) = \frac{1}{2}$.

Thus, Assertion (A) is true.

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Reason (R): $\cos(x + 2n\pi) = \cos x$ for any integer $n$.

This statement is the periodicity property of the cosine function, which states that the value of cosine repeats after every interval of $2\pi$ radians. This property is a fundamental identity in trigonometry.

Thus, Reason (R) is true.

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Explanation:

We calculated $\cos(\frac{17\pi}{3})$ by expressing $\frac{17\pi}{3}$ as $6\pi - \frac{\pi}{3} = (-\frac{\pi}{3}) + 3 \times 2\pi$. Let $x = -\frac{\pi}{3}$ and $n=3$. Then $\frac{17\pi}{3} = x + n \times 2\pi$.

Using the property $\cos(x + 2n\pi) = \cos x$ from Reason (R), we get:

$\cos(\frac{17\pi}{3}) = \cos(-\frac{\pi}{3} + 3 \times 2\pi) = \cos(-\frac{\pi}{3})$.

We then used the identity $\cos(-\theta) = \cos(\theta)$ to get $\cos(-\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$.

The initial step of simplifying the angle $\frac{17\pi}{3}$ by subtracting a multiple of $2\pi$ (specifically $6\pi$) is based directly on the periodicity property stated in Reason (R). Therefore, Reason (R) provides the correct explanation for Assertion (A).

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Since both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A), the correct option is (A).

Given:

$\tan x = \frac{12}{5}$

$x$ lies in the third quadrant.

To Find:

The value of $\text{cosec } x$.

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Solution:

We are given $\tan x = \frac{12}{5}$. We know that $\tan x = \frac{\text{Opposite}}{\text{Adjacent}}$. We can consider a right-angled triangle where the opposite side is 12 and the adjacent side is 5. The hypotenuse can be found using the Pythagorean theorem.

Hypotenuse$^2$ $=$ Opposite$^2$ $+$ Adjacent$^2$

Hypotenuse$^2$ $=$ $12^2 + 5^2$

Hypotenuse$^2$ $=$ $144 + 25$

Hypotenuse$^2$ $=$ $169$

Hypotenuse $=$ $\sqrt{169}$

Hypotenuse $=$ $13$.

Note that the hypotenuse is always positive.

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We need to find the value of $\text{cosec } x$. We know that $\text{cosec } x = \frac{1}{\sin x}$.

In a right-angled triangle, $\sin x = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{12}{13}$.

So, the magnitude of $\sin x$ is $\frac{12}{13}$.

Therefore, the magnitude of $\text{cosec } x$ is $\frac{13}{12}$.

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Now, we need to consider the sign of $\text{cosec } x$. The angle $x$ lies in the third quadrant.

In the third quadrant, the sine function is negative. Since $\text{cosec } x = \frac{1}{\sin x}$, $\text{cosec } x$ is also negative in the third quadrant.

So, $\sin x = -\frac{12}{13}$.

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Therefore, $\text{cosec } x = \frac{1}{\sin x} = \frac{1}{-12/13} = -\frac{13}{12}$.

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This corresponds to option (B).

The final answer is $-\frac{13}{12}$.

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Alternate Solution using Identities:

We are given $\tan x = \frac{12}{5}$. We know the identity $\text{sec}^2 x = 1 + \tan^2 x$.

$\text{sec}^2 x = 1 + \left(\frac{12}{5}\right)^2$

$\text{sec}^2 x = 1 + \frac{144}{25}$

$\text{sec}^2 x = \frac{25 + 144}{25}$

$\text{sec}^2 x = \frac{169}{25}$

$\text{sec } x = \pm \sqrt{\frac{169}{25}} = \pm \frac{13}{5}$.

Since $x$ lies in the third quadrant, $\text{sec } x$ is negative. So, $\text{sec } x = -\frac{13}{5}$.

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We know that $\cos x = \frac{1}{\text{sec } x}$.

$\cos x = \frac{1}{-13/5} = -\frac{5}{13}$.

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We know the identity $\sin^2 x + \cos^2 x = 1$.

$\sin^2 x = 1 - \cos^2 x$

$\sin^2 x = 1 - \left(-\frac{5}{13}\right)^2$

$\sin^2 x = 1 - \frac{25}{169}$

$\sin^2 x = \frac{169 - 25}{169}$

$\sin^2 x = \frac{144}{169}$

$\sin x = \pm \sqrt{\frac{144}{169}} = \pm \frac{12}{13}$.

Since $x$ lies in the third quadrant, $\sin x$ is negative. So, $\sin x = -\frac{12}{13}$.

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Finally, $\text{cosec } x = \frac{1}{\sin x} = \frac{1}{-12/13} = -\frac{13}{12}$.

This matches the previous result.

Let's recall the standard ranges of the given trigonometric functions:

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(i) The range of $\sin x$ is the set of all real numbers from -1 to 1, inclusive. Therefore, the range of $\sin x$ is $[-1, 1]$. This matches options (b) and (d).

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(ii) The range of $\cos x$ is the set of all real numbers from -1 to 1, inclusive. Therefore, the range of $\cos x$ is $[-1, 1]$. This matches options (b) and (d).

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(iii) The range of $\tan x$ is the set of all real numbers. Therefore, the range of $\tan x$ is $(-\infty, \infty)$. This matches option (a).

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(iv) The range of $\sec x$ is the set of all real numbers less than or equal to -1 or greater than or equal to 1. Therefore, the range of $\sec x$ is $(-\infty, -1] \cup [1, \infty)$. This matches option (c).

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Based on the above ranges, the correct matches are:

(i) $\sin x$ $\to$ $[-1, 1]$ (b) or (d)

(ii) $\cos x$ $\to$ $[-1, 1]$ (b) or (d)

(iii) $\tan x$ $\to$ $(-\infty, \infty)$ (a)

(iv) $\sec x$ $\to$ $(-\infty, -1] \cup [1, \infty)$ (c)

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Now let's check the given options:

(A) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)

(i)-(b): $\sin x \to [-1, 1]$ (Correct)

(ii)-(d): $\cos x \to [-1, 1]$ (Correct)

(iii)-(a): $\tan x \to (-\infty, \infty)$ (Correct)

(iv)-(c): $\sec x \to (-\infty, -1] \cup [1, \infty)$ (Correct)

Option (A) provides a correct matching.

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(B) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)

(i)-(d): $\sin x \to [-1, 1]$ (Correct)

(ii)-(b): $\cos x \to [-1, 1]$ (Correct)

(iii)-(a): $\tan x \to (-\infty, \infty)$ (Correct)

(iv)-(c): $\sec x \to (-\infty, -1] \cup [1, \infty)$ (Correct)

Option (B) also provides a correct matching. Note that options (b) and (d) are the same, $[-1, 1]$.

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(C) (i)-(b), (ii)-(b), (iii)-(c), (iv)-(a)

(iii)-(c): $\tan x \to (-\infty, -1] \cup [1, \infty)$ (Incorrect, range of $\tan x$ is $(-\infty, \infty)$)

(iv)-(a): $\sec x \to (-\infty, \infty)$ (Incorrect, range of $\sec x$ is $(-\infty, -1] \cup [1, \infty)$)

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(D) (i)-(d), (ii)-(d), (iii)-(c), (iv)-(a)

(iii)-(c): $\tan x \to (-\infty, -1] \cup [1, \infty)$ (Incorrect)

(iv)-(a): $\sec x \to (-\infty, \infty)$ (Incorrect)

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Both options (A) and (B) represent the correct mapping. Assuming option (A) is the intended answer among the choices provided.

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The final answer is (A).

The domain of the tangent function $\tan y$ is the set of all real numbers $y$ such that $\cos y \neq 0$.

The cosine function $\cos y$ is zero when $y$ is an odd multiple of $\frac{\pi}{2}$.

That is, $y = (2n+1)\frac{\pi}{2}$, where $n$ is any integer ($n \in \mathbb{Z}$).

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For the function $f(x) = \tan(2x)$, the argument of the tangent function is $2x$.

The function $f(x)$ is undefined when $2x$ is an odd multiple of $\frac{\pi}{2}$.

So, we must have $2x \neq (2n+1)\frac{\pi}{2}$ for any integer $n \in \mathbb{Z}$.

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To find the values of $x$ that must be excluded from the domain, we solve for $x$ from the equality:

$2x = (2n+1)\frac{\pi}{2}$

Divide both sides by 2:

$x = \frac{(2n+1)\frac{\pi}{2}}{2}$

$x = (2n+1)\frac{\pi}{4}$

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Therefore, the domain of $f(x) = \tan(2x)$ is the set of all real numbers except those of the form $(2n+1)\frac{\pi}{4}$, where $n$ is any integer.

The domain can be written as $\mathbb{R} - \{(2n+1)\frac{\pi}{4} : n \in \mathbb{Z}\}$.

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This corresponds to option (C).

The final answer is $\mathbb{R} - \{(2n+1)\frac{\pi}{4} : n \in \mathbb{Z}\}$.

To find the value of $\sin 75^\circ$, we can express $75^\circ$ as the sum of two standard angles, such as $45^\circ$ and $30^\circ$.

$75^\circ = 45^\circ + 30^\circ$

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We use the trigonometric identity for the sine of the sum of two angles:

$\sin(A+B) = \sin A \cos B + \cos A \sin B$

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Let $A = 45^\circ$ and $B = 30^\circ$. We know the values of the sine and cosine for these angles:

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\sin 30^\circ = \frac{1}{2}$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

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Substitute these values into the sum formula:

$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$

$\sin 75^\circ = \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{1}{2}\right)$

$\sin 75^\circ = \frac{1 \times \sqrt{3}}{\sqrt{2} \times 2} + \frac{1 \times 1}{\sqrt{2} \times 2}$

$\sin 75^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$

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Combine the fractions since they have a common denominator:

$\sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

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This result matches option (A).

The final answer is $\frac{\sqrt{3}+1}{2\sqrt{2}}$.

We are asked to simplify the expression $\cos(\frac{\pi}{3} + x)\cos x + \sin(\frac{\pi}{3} + x)\sin x$.

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This expression is in the form of the trigonometric identity for the cosine of the difference of two angles, which is:

$\cos(A - B) = \cos A \cos B + \sin A \sin B$

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Comparing the given expression with the identity, we can identify $A$ and $B$.

Let $A = \frac{\pi}{3} + x$ and $B = x$.

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Substitute these values into the identity $\cos(A - B)$:

$\cos(A - B) = \cos\left(\left(\frac{\pi}{3} + x\right) - x\right)$

$\cos(A - B) = \cos\left(\frac{\pi}{3} + x - x\right)$

$\cos(A - B) = \cos\left(\frac{\pi}{3}\right)$

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Thus, the simplified expression is $\cos(\frac{\pi}{3})$.

The value of $\cos(\frac{\pi}{3})$ is $\frac{1}{2}$. However, the options are given in terms of trigonometric functions of $\frac{\pi}{3}$ or expressions involving $x$.

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The simplified form $\cos(\frac{\pi}{3})$ matches option (B).

The final answer is $\cos(\frac{\pi}{3})$.

Given:

$\sin A = \frac{4}{5}$

$\cos B = \frac{12}{13}$

Angles A and B are in the first quadrant.

To Find:

The value of $\cos(A-B)$.

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Solution:

We need to find the value of $\cos(A-B)$ using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$.

We are given $\sin A = \frac{4}{5}$ and $\cos B = \frac{12}{13}$. We need to find $\cos A$ and $\sin B$.

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For angle A, which is in the first quadrant, $\cos A$ is positive. We use the identity $\sin^2 A + \cos^2 A = 1$.

$\cos^2 A = 1 - \sin^2 A$

$\cos^2 A = 1 - \left(\frac{4}{5}\right)^2$

$\cos^2 A = 1 - \frac{16}{25}$

$\cos^2 A = \frac{25 - 16}{25}$

$\cos^2 A = \frac{9}{25}$

Since A is in the first quadrant, $\cos A = \sqrt{\frac{9}{25}} = \frac{3}{5}$.

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For angle B, which is in the first quadrant, $\sin B$ is positive. We use the identity $\sin^2 B + \cos^2 B = 1$.

$\sin^2 B = 1 - \cos^2 B$

$\sin^2 B = 1 - \left(\frac{12}{13}\right)^2$

$\sin^2 B = 1 - \frac{144}{169}$

$\sin^2 B = \frac{169 - 144}{169}$

$\sin^2 B = \frac{25}{169}$

Since B is in the first quadrant, $\sin B = \sqrt{\frac{25}{169}} = \frac{5}{13}$.

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Now, substitute the values of $\sin A$, $\cos A$, $\sin B$, and $\cos B$ into the formula for $\cos(A-B)$:

$\cos(A-B) = \cos A \cos B + \sin A \sin B$

$\cos(A-B) = \left(\frac{3}{5}\right) \times \left(\frac{12}{13}\right) + \left(\frac{4}{5}\right) \times \left(\frac{5}{13}\right)$

$\cos(A-B) = \frac{3 \times 12}{5 \times 13} + \frac{4 \times 5}{5 \times 13}$

$\cos(A-B) = \frac{36}{65} + \frac{20}{65}$

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Add the fractions:

$\cos(A-B) = \frac{36 + 20}{65}$

$\cos(A-B) = \frac{56}{65}$

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This result matches option (A).

The final answer is $\frac{56}{65}$.

We are asked to find the general solution of the equation $\sin x = \frac{1}{2}$.

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First, we find the principal value of $x$ for which $\sin x = \frac{1}{2}$. We know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$.

So, the equation can be written as:

$\sin x = \sin(\frac{\pi}{6})$

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The general solution for trigonometric equations of the form $\sin x = \sin y$ is given by the formula:

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Here, $y = \frac{\pi}{6}$. Substituting this value into the general solution formula, we get:

$x = n\pi + (-1)^n \frac{\pi}{6}$

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This is the general solution for the equation $\sin x = \frac{1}{2}$.

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This matches option (A).

The final answer is $x = n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbb{Z}$.

We are asked to find the general solution of the equation $\tan x = -1$.

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First, we find a principal value of $x$ for which $\tan x = -1$.

We know that $\tan(\frac{\pi}{4}) = 1$. Since the tangent function is negative in the second and fourth quadrants, we can find an angle in either quadrant that has a tangent of -1.

Using the property $\tan(\pi - \theta) = -\tan \theta$, we have $\tan(\pi - \frac{\pi}{4}) = \tan(\frac{3\pi}{4}) = -\tan(\frac{\pi}{4}) = -1$. So, $\frac{3\pi}{4}$ is a value for which $\tan x = -1$.

Alternatively, using the property $\tan(-\theta) = -\tan \theta$, we have $\tan(-\frac{\pi}{4}) = -\tan(\frac{\pi}{4}) = -1$. So, $-\frac{\pi}{4}$ is another valid value.

Let's use $y = -\frac{\pi}{4}$. The equation is $\tan x = \tan(-\frac{\pi}{4})$.

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The general solution for a trigonometric equation of the form $\tan x = \tan y$ is given by:

$x = n\pi + y$, where $n \in \mathbb{Z}$

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Substitute $y = -\frac{\pi}{4}$ into the general solution formula:

$x = n\pi + (-\frac{\pi}{4})$

$x = n\pi - \frac{\pi}{4}$

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This is the general solution for the equation $\tan x = -1$.

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This matches option (B).

The final answer is $x = n\pi - \frac{\pi}{4}, n \in \mathbb{Z}$.

We need to identify the correct trigonometric identities among the given options.

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Let's examine each option:

(A) $\sin^2 \theta + \cos^2 \theta = 1$

This is the fundamental Pythagorean trigonometric identity, which is true for all real values of $\theta$.

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(B) $\sec^2 \theta - \tan^2 \theta = 1$

This is another standard trigonometric identity, derived from dividing the fundamental identity by $\cos^2 \theta$. It is true for all real values of $\theta$ where $\cos \theta \neq 0$.

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(C) $\text{cosec}^2 \theta - \cot^2 \theta = 1$

This is also a standard trigonometric identity, derived from dividing the fundamental identity by $\sin^2 \theta$. It is true for all real values of $\theta$ where $\sin \theta \neq 0$.

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(D) All of the above

Since the identities given in options (A), (B), and (C) are all correct standard trigonometric identities, the statement "All of the above" is correct.

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Therefore, all the listed options (A), (B), and (C) are correct identities.

The correct choice is the one that encompasses all the correct individual options.

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The final answer is (D) All of the above.

We need to determine which of the given options is not the range of the trigonometric function $\sec(x)$.

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The function $\sec(x)$ is defined as $\sec(x) = \frac{1}{\cos(x)}$.

The range of $\cos(x)$ is known to be the interval $[-1, 1]$. That is, $-1 \leq \cos(x) \leq 1$ for all real values of $x$.

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Now, let's consider the values that $\sec(x) = \frac{1}{\cos(x)}$ can take:

If $\cos(x) > 0$, then $0 < \cos(x) \leq 1$. Taking the reciprocal, we get $\frac{1}{\cos(x)} \geq \frac{1}{1}$, which means $\sec(x) \geq 1$. Since $\cos(x)$ can be arbitrarily close to 0 (from the positive side), $\sec(x)$ can be arbitrarily large. So, when $\cos(x) > 0$, the values of $\sec(x)$ are in the interval $[1, \infty)$.

If $\cos(x) < 0$, then $-1 \leq \cos(x) < 0$. Taking the reciprocal and reversing the inequality sign (because we are dealing with negative numbers), we get $\frac{1}{\cos(x)} \leq \frac{1}{-1}$, which means $\sec(x) \leq -1$. Since $\cos(x)$ can be arbitrarily close to 0 (from the negative side), $\sec(x)$ can be arbitrarily small (large in negative magnitude). So, when $\cos(x) < 0$, the values of $\sec(x)$ are in the interval $(-\infty, -1]$.

---

The value $\cos(x)=0$ is excluded from the domain of $\sec(x)$, as division by zero is undefined.

---

Combining these two cases, the range of $\sec(x)$ is the union of these two intervals:

---

Now, let's compare this range with the given options:

• (A) $(-\infty, -1]$ is a part of the range of $\sec(x)$.
• (B) $[1, \infty)$ is a part of the range of $\sec(x)$.
• (C) $[-1, 1]$ is not the range of $\sec(x)$. In fact, this is the interval of values that $\sec(x)$ can never take.
• (D) $(-\infty, -1] \cup [1, \infty)$ is the complete range of $\sec(x)$.

---

The question asks which of the following is NOT the range of $\sec(x)$. Based on our analysis, the interval $[-1, 1]$ is the one that is not the range (or any part of the range other than the endpoints, which are included in the range of $\sec(x)$ as part of the intervals $(-\infty, -1]$ and $[1, \infty)$). The interval $(-1, 1)$ is completely outside the range of $\sec(x)$. Option (C) represents the interval $[-1, 1]$, which is precisely the set of values that $\sec(x)$ does not attain (except possibly at the endpoints, but the interval itself represents values between -1 and 1 inclusive). Thus, $[-1, 1]$ is not the range of $\sec(x)$.

---

The final answer is (C) $[-1, 1]$.

We need to find the value of the expression $\sin(\frac{\pi}{2} - x)$.

---

This expression involves the sine of a difference of angles, specifically a difference involving $\frac{\pi}{2}$.

We can use the trigonometric identity for the sine of the difference of two angles:

$\sin(A - B) = \sin A \cos B - \cos A \sin B$

---

In this case, we have $A = \frac{\pi}{2}$ and $B = x$.

Substituting these values into the identity, we get:

$\sin(\frac{\pi}{2} - x) = \sin(\frac{\pi}{2}) \cos x - \cos(\frac{\pi}{2}) \sin x$

---

We know the values of $\sin(\frac{\pi}{2})$ and $\cos(\frac{\pi}{2})$:

---

Substitute these values back into the expression:

$\sin(\frac{\pi}{2} - x) = (1) \cdot \cos x - (0) \cdot \sin x$

$\sin(\frac{\pi}{2} - x) = \cos x - 0$

$\sin(\frac{\pi}{2} - x) = \cos x$

---

This is a fundamental complementary angle identity. The sine of an angle is equal to the cosine of its complementary angle.

---

Comparing our result with the given options, we find that the value of $\sin(\frac{\pi}{2} - x)$ is $\cos x$, which corresponds to option (C).

---

The final answer is (C) $\cos x$.

Given:

The company's quarterly profit (in lakhs of $\textsf{₹}$) is modelled by the function:

where $q = 1, 2, 3, 4$ represent the quarters (Q1, Q2, Q3, Q4).

---

To Find:

The maximum profit the company can achieve in a quarter based on this model.

---

Solution:

The profit function is given by $P(q) = 10 + 5\sin(\frac{\pi}{2} q)$.

To find the maximum profit, we need to find the maximum value of the function $P(q)$ for $q \in \{1, 2, 3, 4\}$.

The value of $P(q)$ depends on the term $\sin(\frac{\pi}{2} q)$.

We know that the range of the sine function for any real angle is $[-1, 1]$.

This means that for any value of $q$, the value of $\sin(\frac{\pi}{2} q)$ is between -1 and 1, inclusive.

---

Now, let's use this inequality to find the range of $P(q)$.

Multiply the inequality by 5 (which is a positive number):

---

Now, add 10 to all parts of the inequality:

This inequality shows that the profit $P(q)$ ranges from a minimum value of 5 to a maximum value of 15.

---

The maximum value of $P(q)$ is 15.

This maximum profit is achieved when $\sin(\frac{\pi}{2} q)$ reaches its maximum value, which is 1.

For the given values of $q \in \{1, 2, 3, 4\}$:

• For $q=1$, $\sin(\frac{\pi}{2} \times 1) = \sin(\frac{\pi}{2}) = 1$. Profit = $10 + 5(1) = 15$.
• For $q=2$, $\sin(\frac{\pi}{2} \times 2) = \sin(\pi) = 0$. Profit = $10 + 5(0) = 10$.
• For $q=3$, $\sin(\frac{\pi}{2} \times 3) = \sin(\frac{3\pi}{2}) = -1$. Profit = $10 + 5(-1) = 5$.
• For $q=4$, $\sin(\frac{\pi}{2} \times 4) = \sin(2\pi) = 0$. Profit = $10 + 5(0) = 10$.

The maximum profit observed for $q \in \{1, 2, 3, 4\}$ is $\textsf{₹} 15$ lakhs.

---

The final answer is (B) $\textsf{₹} 15$ lakhs.

Given:

The company's quarterly profit (in lakhs of $\textsf{₹}$) is modelled by the function:

where $q = 1, 2, 3, 4$ represent the quarters (Q1, Q2, Q3, Q4).

---

To Find:

The minimum profit the company can achieve in a quarter based on this model.

---

Solution:

The profit function is given by $P(q) = 10 + 5\sin(\frac{\pi}{2} q)$.

To find the minimum profit, we need to find the minimum value of the function $P(q)$ for $q \in \{1, 2, 3, 4\}$.

The value of $P(q)$ depends on the term $\sin(\frac{\pi}{2} q)$.

We know that the range of the sine function for any real angle is $[-1, 1]$.

This means that for any value of $q$, the value of $\sin(\frac{\pi}{2} q)$ is between -1 and 1, inclusive.

---

Now, let's use this inequality to find the range of $P(q)$.

Multiply the inequality by 5 (which is a positive number):

---

Now, add 10 to all parts of the inequality:

This inequality shows that the profit $P(q)$ ranges from a minimum value of 5 to a maximum value of 15.

---

The minimum value of $P(q)$ is 5.

This minimum profit is achieved when $\sin(\frac{\pi}{2} q)$ reaches its minimum value, which is -1.

For the given values of $q \in \{1, 2, 3, 4\}$:

• For $q=1$, $\sin(\frac{\pi}{2} \times 1) = \sin(\frac{\pi}{2}) = 1$. Profit = $10 + 5(1) = 15$.
• For $q=2$, $\sin(\frac{\pi}{2} \times 2) = \sin(\pi) = 0$. Profit = $10 + 5(0) = 10$.
• For $q=3$, $\sin(\frac{\pi}{2} \times 3) = \sin(\frac{3\pi}{2}) = -1$. Profit = $10 + 5(-1) = 5$.
• For $q=4$, $\sin(\frac{\pi}{2} \times 4) = \sin(2\pi) = 0$. Profit = $10 + 5(0) = 10$.

The minimum profit observed for $q \in \{1, 2, 3, 4\}$ is $\textsf{₹} 5$ lakhs.

---

The final answer is (C) $\textsf{₹} 5$ lakhs.

Given:

The company's quarterly profit (in lakhs of $\textsf{₹}$) is modelled by the function:

where $q = 1, 2, 3, 4$ represent the quarters (Q1, Q2, Q3, Q4).

---

To Find:

The predicted profit for the 3rd quarter ($q=3$).

---

Solution:

To find the predicted profit for the 3rd quarter, we substitute $q=3$ into the profit function $P(q)$.

$P(3) = 10 + 5\sin(\frac{\pi}{2} \times 3)$

$P(3) = 10 + 5\sin(\frac{3\pi}{2})$

---

We need to find the value of $\sin(\frac{3\pi}{2})$.

The angle $\frac{3\pi}{2}$ radians is equivalent to $270^\circ$.

The sine of $\frac{3\pi}{2}$ is -1.

---

Now substitute this value back into the expression for $P(3)$:

$P(3) = 10 + 5(-1)$

$P(3) = 10 - 5$

$P(3) = 5$

---

So, the predicted profit for the 3rd quarter is 5 lakhs of $\textsf{₹}$.

---

Comparing our result with the given options, we find that the profit is $\textsf{₹} 5$ lakhs, which corresponds to option (C).

---

The final answer is (C) $\textsf{₹} 5$ lakhs.

We need to identify the correct description of the graph of the function $y = \cos x$.

---

Let's recall the properties of the cosine function and its graph.

The cosine function is defined for all real numbers.

The range of the cosine function is $[-1, 1]$.

The cosine function is periodic with a period of $2\pi$. This means $\cos(x + 2\pi) = \cos x$ for all $x$.

---

Let's check the value of $y = \cos x$ at some key points:

• At $x=0$, $y = \cos(0) = 1$. The graph starts at the point $(0, 1)$.
• At $x=\frac{\pi}{2}$, $y = \cos(\frac{\pi}{2}) = 0$. The graph passes through the point $(\frac{\pi}{2}, 0)$.
• At $x=\pi$, $y = \cos(\pi) = -1$. The graph passes through the point $(\pi, -1)$.
• At $x=\frac{3\pi}{2}$, $y = \cos(\frac{3\pi}{2}) = 0$. The graph passes through the point $(\frac{3\pi}{2}, 0)$.
• At $x=2\pi$, $y = \cos(2\pi) = 1$. The graph passes through the point $(2\pi, 1)$, completing one cycle.

---

Now let's evaluate the given options:

(A) Starts at (0,1), passes through $(\pi/2, 0)$, $(\pi, -1)$, etc., periodic with period $2\pi$.

This option correctly describes the starting point $(0, 1)$, passes through the indicated points, and states the correct period ($2\pi$).

---

(B) Starts at (0,0), passes through $(\pi/2, 1)$, $(\pi, 0)$, etc., periodic with period $2\pi$.

The graph of $y = \cos x$ starts at $(0, 1)$, not $(0, 0)$. This option describes the graph of $y = \sin x$. Therefore, this option is incorrect.

---

(C) Has vertical asymptotes at $x = (n + 1/2)\pi$, $n \in \mathbb{Z}$, periodic with period $\pi$.

The graph of $y = \cos x$ does not have vertical asymptotes because the cosine function is defined for all real numbers. Functions like $\tan x$ and $\sec x$ have vertical asymptotes. Also, the period is $2\pi$, not $\pi$. This option is incorrect.

---

(D) Has vertical asymptotes at $x = n\pi$, $n \in \mathbb{Z}$, periodic with period $\pi$.

Similar to option (C), the graph of $y = \cos x$ does not have vertical asymptotes. Functions like $\cot x$ and $\text{cosec} x$ have vertical asymptotes at $x = n\pi$. Also, the period is $2\pi$, not $\pi$. This option is incorrect.

---

Based on the analysis, option (A) provides the correct description of the graph of $y = \cos x$.

---

The final answer is (A) Starts at (0,1), passes through $(\pi/2, 0)$, $(\pi, -1)$, etc., periodic with period $2\pi$.

Given:

The function is $f(x) = \sin(x/3)$.

---

To Find:

The period of the function $f(x)$.

---

Solution:

The general form of a sinusoidal function is $g(x) = A \sin(Bx + C) + D$ or $g(x) = A \cos(Bx + C) + D$.

The period of such a function is given by the formula:

---

Our given function is $f(x) = \sin(x/3)$.

We can rewrite this as $f(x) = \sin(\frac{1}{3}x)$.

Comparing this with the form $\sin(Bx)$, we can identify the value of $B$.

---

Now, we substitute the value of $B$ into the period formula (i):

Period $= \frac{2\pi}{|1/3|}$

Period $= \frac{2\pi}{1/3}$

Period $= 2\pi \times 3$

---

Thus, the period of the function $f(x) = \sin(x/3)$ is $6\pi$.

---

Comparing our result with the given options, we find that the period is $6\pi$, which corresponds to option (C).

---

The final answer is (C) $6\pi$.

Given:

The function is $f(x) = 3\cos x + 2$.

---

To Find:

The range of the function $f(x)$.

---

Solution:

To find the range of $f(x) = 3\cos x + 2$, we use the known range of the cosine function.

The range of $\cos x$ for all real values of $x$ is the interval $[-1, 1]$.

---

Next, we apply the transformations to this inequality to find the range of $f(x)$.

First, multiply the inequality by the coefficient of $\cos x$, which is 3.

Multiplying by a positive number does not change the direction of the inequality signs.

---

Finally, we add the constant term, 2, to all parts of the inequality.

Adding a number to all parts of an inequality does not change the direction of the inequality signs.

---

Since $f(x) = 3\cos x + 2$, the inequality (iii) tells us that the value of $f(x)$ is between -1 and 5, inclusive.

Therefore, the range of the function $f(x) = 3\cos x + 2$ is the interval $[-1, 5]$.

---

The final answer is (B) $[1, 5]$.

Given:

The trigonometric equation is $\sqrt{3} \tan x = -1$.

We need to find the principal solutions for $x$ in the interval $[0, 2\pi)$.

---

To Find:

The value(s) of $x \in [0, 2\pi)$ that satisfy the given equation.

---

Solution:

First, let's isolate $\tan x$ in the given equation:

Divide both sides by $\sqrt{3}$:

---

We know that $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.

Since $\tan x$ is negative, the angle $x$ must lie in the second or fourth quadrant, as the tangent function is negative in these quadrants.

---

For the second quadrant, the angle is of the form $\pi - \theta$, where $\tan \theta = \frac{1}{\sqrt{3}}$, so $\theta = \frac{\pi}{6}$.

The angle in the second quadrant is $\pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}$.

Let's check if $\tan(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$:

$\tan(\frac{5\pi}{6}) = -\tan(\frac{\pi}{6})$

$\tan(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$

So, $x = \frac{5\pi}{6}$ is a solution.

This value is in the interval $[0, 2\pi)$ since $0 \leq \frac{5\pi}{6} < 2\pi$.

---

For the fourth quadrant, the angle is of the form $2\pi - \theta$, where $\tan \theta = \frac{1}{\sqrt{3}}$, so $\theta = \frac{\pi}{6}$.

The angle in the fourth quadrant is $2\pi - \frac{\pi}{6} = \frac{12\pi - \pi}{6} = \frac{11\pi}{6}$.

Let's check if $\tan(\frac{11\pi}{6}) = -\frac{1}{\sqrt{3}}$:

$\tan(\frac{11\pi}{6}) = -\tan(\frac{\pi}{6})$

$\tan(\frac{11\pi}{6}) = -\frac{1}{\sqrt{3}}$

So, $x = \frac{11\pi}{6}$ is another solution.

This value is also in the interval $[0, 2\pi)$ since $0 \leq \frac{11\pi}{6} < 2\pi$.

---

The principal solutions in the interval $[0, 2\pi)$ are $x = \frac{5\pi}{6}$ and $x = \frac{11\pi}{6}$.

---

Comparing these solutions with the given options, we find that option (B) lists both $\frac{5\pi}{6}$ and $\frac{11\pi}{6}$.

---

The final answer is (B) $\frac{5\pi}{6}, \frac{11\pi}{6}$.

Given:

The trigonometric equation $\cos x = \cos y$.

---

To Find:

The general solution for $x$ in terms of $y$, where $n$ is an integer.

---

Solution:

We are given the equation:

$\cos x = \cos y$

Rearranging the terms, we get:

$\cos x - \cos y = 0$

---

We use the sum-to-product trigonometric identity:

$\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$

Applying this identity to our equation with $A=x$ and $B=y$, we have:

---

For the product of two terms to be zero, at least one of the terms must be zero.

Thus, equation (i) is true if and only if $\sin\left(\frac{x+y}{2}\right) = 0$ or $\sin\left(\frac{x-y}{2}\right) = 0$.

---

Case 1: $\sin\left(\frac{x+y}{2}\right) = 0$

The general solution for $\sin \theta = 0$ is $\theta = n\pi$, where $n$ is any integer ($n \in \mathbb{Z}$).

So, $\frac{x+y}{2} = n\pi$ for some integer $n$.

Multiplying both sides by 2, we get:

$x+y = 2n\pi$

Solving for $x$:

$x = 2n\pi - y$, where $n \in \mathbb{Z}$.

---

Case 2: $\sin\left(\frac{x-y}{2}\right) = 0$

Using the same general solution for $\sin \theta = 0$, we have:

$\frac{x-y}{2} = m\pi$ for some integer $m$ ($m \in \mathbb{Z}$).

Multiplying both sides by 2, we get:

$x-y = 2m\pi$

Solving for $x$:

$x = 2m\pi + y$, where $m \in \mathbb{Z}$.

---

Combining the solutions from Case 1 and Case 2, we find that $x$ must be of the form $2 \times (\text{integer}) \times \pi \pm y$. Since both $n$ and $m$ can be any integer, we can express the general solution using a single integer $n$ (or any other variable like $k$) as:

---

Comparing this general solution with the given options:

Option (A) $x = n\pi + y$ is incorrect because it includes terms like $\pi+y, 3\pi+y$, etc., where $\cos(\pi+y) = -\cos y$, not necessarily $\cos y$.

Option (B) $x = n\pi \pm y$ is incorrect because it includes odd multiples of $\pi$, like $\pi \pm y$. $\cos(\pi \pm y) = -\cos y$, not necessarily $\cos y$.

Option (C) $x = 2n\pi + y$ only represents the case where $x-y$ is a multiple of $2\pi$. It misses the case where $x+y$ is a multiple of $2\pi$.

Option (D) $x = 2n\pi \pm y$ correctly represents both cases derived from the identity.

---

The final answer is (D) $x = 2n\pi \pm y, n \in \mathbb{Z}$.

We are given an Assertion (A) and a Reason (R) and need to determine their truthfulness and if the Reason correctly explains the Assertion.

---

Assertion (A): The range of $\text{cosec } x$ is $\mathbb{R} - (-1, 1)$.

The range of the cosecant function, $\text{cosec } x$, is the set of all real numbers except those in the open interval $(-1, 1)$. This is typically written as $(-\infty, -1] \cup [1, \infty)$. The set $\mathbb{R} - (-1, 1)$ is equivalent to $(-\infty, -1] \cup [1, \infty)$.

So, Assertion (A) is True.

---

Reason (R): $\text{cosec } x = 1/\sin x$, and the range of $\sin x$ is $[-1, 1]$.

The definition of the cosecant function is indeed $\text{cosec } x = \frac{1}{\sin x}$, provided $\sin x \neq 0$.

The range of the sine function, $\sin x$, is the closed interval $[-1, 1]$, meaning that for any real number $x$, $-1 \leq \sin x \leq 1$.

So, Reason (R) is True.

---

Now, we need to check if Reason (R) is the correct explanation for Assertion (A).

The function $\text{cosec } x = \frac{1}{\sin x}$. The values of $\text{cosec } x$ are the reciprocals of the values of $\sin x$.

The range of $\sin x$ is $[-1, 1]$. This means the values of $\sin x$ are between -1 and 1, inclusive.

• If $\sin x = 1$, then $\text{cosec } x = \frac{1}{1} = 1$.
• If $\sin x = -1$, then $\text{cosec } x = \frac{1}{-1} = -1$.
• If $\sin x$ is in $(0, 1]$, then $\frac{1}{\sin x}$ is in $[1, \infty)$. For example, if $\sin x = 0.5$, $\text{cosec } x = 2$; if $\sin x = 0.01$, $\text{cosec } x = 100$.
• If $\sin x$ is in $[-1, 0)$, then $\frac{1}{\sin x}$ is in $(-\infty, -1]$. For example, if $\sin x = -0.5$, $\text{cosec } x = -2$; if $\sin x = -0.01$, $\text{cosec } x = -100$.
• The value $\sin x = 0$ is excluded from the domain of $\text{cosec } x$, as $\text{cosec } x$ would be undefined.

Combining these cases, the values of $\text{cosec } x$ can be any number in $(-\infty, -1] \cup [1, \infty)$. The values in the interval $(-1, 1)$ are not attained by $\text{cosec } x$ because if $|\sin x| \leq 1$, then $|\frac{1}{\sin x}| \geq 1$ (when $\sin x \neq 0$).

Thus, the range of $\text{cosec } x$ is indeed $(-\infty, -1] \cup [1, \infty)$, which is $\mathbb{R} - (-1, 1)$.

The reason given explains this fact by stating the definition of $\text{cosec } x$ and the range of $\sin x$, from which the range of $\text{cosec } x$ is derived.

Therefore, Reason (R) is the correct explanation of Assertion (A).

---

Since both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A), the correct option is (A).

---

The final answer is (A) Both A and R are true and R is the correct explanation of A.

Given:

We need to find the value of $\tan(105^\circ)$.

---

Solution:

We can express $105^\circ$ as the sum of two standard angles for which we know the tangent values. A common choice is $60^\circ$ and $45^\circ$.

$105^\circ = 60^\circ + 45^\circ$

---

We use the tangent addition formula:

$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

---

Substitute $A = 60^\circ$ and $B = 45^\circ$ into the formula:

$\tan(105^\circ) = \tan(60^\circ + 45^\circ) = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ}$

---

We know the exact values of $\tan 60^\circ$ and $\tan 45^\circ$:

$\tan 60^\circ = \sqrt{3}$

$\tan 45^\circ = 1$

---

Substitute these values into the expression:

$\tan(105^\circ) = \frac{\sqrt{3} + 1}{1 - (\sqrt{3})(1)}$

$\tan(105^\circ) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$

---

To simplify this expression and remove the radical from the denominator, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of $1 - \sqrt{3}$, which is $1 + \sqrt{3}$.

$\tan(105^\circ) = \frac{(\sqrt{3} + 1)}{(1 - \sqrt{3})} \times \frac{(1 + \sqrt{3})}{(1 + \sqrt{3})}$

Now, we multiply the terms in the numerator and the denominator.

Numerator: $(\sqrt{3} + 1)(1 + \sqrt{3}) = (\sqrt{3})(1) + (\sqrt{3})(\sqrt{3}) + (1)(1) + (1)(\sqrt{3}) = \sqrt{3} + 3 + 1 + \sqrt{3} = 4 + 2\sqrt{3}$

Denominator: $(1 - \sqrt{3})(1 + \sqrt{3})$. Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$, we get $1^2 - (\sqrt{3})^2 = 1 - 3 = -2$.

---

So, the expression becomes:

$\tan(105^\circ) = \frac{4 + 2\sqrt{3}}{-2}$

Now, divide the numerator by -2:

$\tan(105^\circ) = \frac{2(2 + \sqrt{3})}{-2}$

$\tan(105^\circ) = -(2 + \sqrt{3})$

$\tan(105^\circ) = -2 - \sqrt{3}$

---

Alternate Solution:

We can also express $105^\circ$ as the difference of two standard angles, such as $150^\circ$ and $45^\circ$.

$105^\circ = 150^\circ - 45^\circ$

---

We use the tangent subtraction formula:

$\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

---

Substitute $A = 150^\circ$ and $B = 45^\circ$. We know $\tan 150^\circ = \tan(180^\circ - 30^\circ) = -\tan 30^\circ = -\frac{1}{\sqrt{3}}$ and $\tan 45^\circ = 1$.

$\tan(105^\circ) = \frac{-\frac{1}{\sqrt{3}} - 1}{1 + (-\frac{1}{\sqrt{3}})(1)} = \frac{-\frac{1}{\sqrt{3}} - 1}{1 - \frac{1}{\sqrt{3}}}$

---

Simplify the complex fraction by finding a common denominator in the numerator and denominator:

$\tan(105^\circ) = \frac{\frac{-1 - \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}}$

$\tan(105^\circ) = \frac{-1 - \sqrt{3}}{\sqrt{3} - 1}$

Now, rationalize the denominator by multiplying the numerator and denominator by the conjugate of $\sqrt{3} - 1$, which is $\sqrt{3} + 1$.

$\tan(105^\circ) = \frac{(-1 - \sqrt{3})}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$

Numerator: $(-1 - \sqrt{3})(\sqrt{3} + 1) = -1(\sqrt{3}) -1(1) - \sqrt{3}(\sqrt{3}) - \sqrt{3}(1) = -\sqrt{3} - 1 - 3 - \sqrt{3} = -4 - 2\sqrt{3}$. Alternatively, $-(\sqrt{3}+1)(\sqrt{3}+1) = -(\sqrt{3}+1)^2 = -((\sqrt{3})^2 + 2\sqrt{3} + 1^2) = -(3 + 2\sqrt{3} + 1) = -(4 + 2\sqrt{3})$.

Denominator: $(\sqrt{3} - 1)(\sqrt{3} + 1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2$.

---

So, the expression becomes:

$\tan(105^\circ) = \frac{-4 - 2\sqrt{3}}{2}$

Divide the numerator by 2:

$\tan(105^\circ) = \frac{-2(2 + \sqrt{3})}{2}$

$\tan(105^\circ) = -(2 + \sqrt{3})$

$\tan(105^\circ) = -2 - \sqrt{3}$

---

Both methods yield the same result.

Comparing our result with the given options, we find that the value of $\tan(105^\circ)$ is $-2 - \sqrt{3}$, which corresponds to option (D).

---

The final answer is (D) $-2 - \sqrt{3}$.

Given:

We are given that $\sin x = \frac{1}{3}$ and that the angle $x$ lies in the second quadrant.

---

To Find:

The value of $\cos x$.

---

Solution:

We can use the fundamental trigonometric identity that relates $\sin x$ and $\cos x$:

---

Substitute the given value of $\sin x = \frac{1}{3}$ into the identity:

$(\frac{1}{3})^2 + \cos^2 x = 1$

$\frac{1}{9} + \cos^2 x = 1$

---

Now, solve for $\cos^2 x$:

$\cos^2 x = 1 - \frac{1}{9}$

To subtract the fractions, find a common denominator, which is 9:

$\cos^2 x = \frac{9}{9} - \frac{1}{9}$

---

Now, take the square root of both sides to find $\cos x$:

$\cos x = \pm \sqrt{\frac{8}{9}}$

$\cos x = \pm \frac{\sqrt{8}}{\sqrt{9}}$

$\cos x = \pm \frac{\sqrt{4 \times 2}}{3}$

---

We are given that $x$ is in the second quadrant.

In the second quadrant, the values of cosine are negative.

---

Therefore, we must choose the negative sign from the possible values in equation (iii).

$\cos x = -\frac{2\sqrt{2}}{3}$

---

Comparing this value with the given options:

• (A) $\frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$ (Positive)
• (B) $-\frac{\sqrt{8}}{3} = -\frac{2\sqrt{2}}{3}$ (Negative)
• (C) $\frac{2\sqrt{2}}{3}$ (Positive)
• (D) $-\frac{2\sqrt{2}}{3}$ (Negative)

Both option (B) and (D) represent the same negative value, but option (D) presents the value with $\sqrt{8}$ simplified to $2\sqrt{2}$. Option (D) matches our calculated value exactly.

---

The final answer is (D) $-\frac{2\sqrt{2}}{3}$.

Given:

The graph of the function $y = \sin x$ starts at $(0, 0)$ and increases to a maximum at $x = \pi/2$.

---

To Find:

The next point where the graph of $y = \sin x$ crosses the x-axis after $x=0$ in the positive direction.

---

Solution:

The graph of a function crosses the x-axis when the value of the function is zero, i.e., $y = 0$.

For the function $y = \sin x$, we need to find the values of $x$ for which $\sin x = 0$.

---

The general solution for the equation $\sin x = 0$ is given by:

---

This means the graph of $y = \sin x$ crosses the x-axis at integer multiples of $\pi$.

Let's list some of these x-intercepts for integer values of $n$:

• If $n=0$, $x = 0 \times \pi = 0$. The point is $(0, 0)$. This is the starting point mentioned in the question.
• If $n=1$, $x = 1 \times \pi = \pi$. The point is $(\pi, 0)$.
• If $n=2$, $x = 2 \times \pi = 2\pi$. The point is $(2\pi, 0)$.
• If $n=-1$, $x = -1 \times \pi = -\pi$. The point is $(-\pi, 0)$.

---

The question states that the graph starts at $(0, 0)$ and increases to a maximum at $x = \pi/2$. This describes the behaviour of the sine function for $x$ values from 0 to $\pi/2$.

The first x-intercept is at $x=0$. We are looking for the *next* point where it crosses the x-axis for $x > 0$.

Looking at the list of positive x-intercepts ($0, \pi, 2\pi, \dots$), the next value after 0 is $\pi$.

The point where the graph crosses the x-axis after $(0,0)$ is $(\pi, 0)$.

---

Let's check the options:

• (A) $\pi/2$: At $x = \pi/2$, $y = \sin(\pi/2) = 1$. This is the maximum point, not an x-intercept.
• (B) $\pi$: At $x = \pi$, $y = \sin(\pi) = 0$. This is an x-intercept. It is the next one after $x=0$ in the positive direction.
• (C) $3\pi/2$: At $x = 3\pi/2$, $y = \sin(3\pi/2) = -1$. This is the minimum point, not an x-intercept.
• (D) $2\pi$: At $x = 2\pi$, $y = \sin(2\pi) = 0$. This is an x-intercept, but it is the one after $\pi$.

---

The next point where the graph crosses the x-axis after $(0,0)$ is at $x=\pi$.

---

The final answer is (B) $\pi$.

Given:

We are given that $\sin A = \frac{3}{5}$ and $\cos B = \frac{5}{13}$, where A and B are acute angles.

---

To Find:

The value of $\sin(A+B)$.

---

Solution:

We use the trigonometric identity for the sine of the sum of two angles:

$\sin(A+B) = \sin A \cos B + \cos A \sin B$

To use this formula, we need the values of $\sin A$, $\cos B$, $\cos A$, and $\sin B$.

We are given $\sin A = \frac{3}{5}$ and $\cos B = \frac{5}{13}$.

Since A and B are acute angles (i.e., in the first quadrant), both $\sin$ and $\cos$ values are positive for angles A and B.

---

We can find $\cos A$ using the identity $\sin^2 A + \cos^2 A = 1$:

$\cos^2 A = 1 - \sin^2 A$

$\cos^2 A = 1 - \left(\frac{3}{5}\right)^2$

$\cos^2 A = 1 - \frac{9}{25}$

$\cos^2 A = \frac{25 - 9}{25}$

$\cos^2 A = \frac{16}{25}$

Since A is acute, $\cos A$ is positive.

$\cos A = \sqrt{\frac{16}{25}} = \frac{4}{5}$

---

Similarly, we can find $\sin B$ using the identity $\sin^2 B + \cos^2 B = 1$:

$\sin^2 B = 1 - \cos^2 B$

$\sin^2 B = 1 - \left(\frac{5}{13}\right)^2$

$\sin^2 B = 1 - \frac{25}{169}$

$\sin^2 B = \frac{169 - 25}{169}$

$\sin^2 B = \frac{144}{169}$

Since B is acute, $\sin B$ is positive.

$\sin B = \sqrt{\frac{144}{169}} = \frac{12}{13}$

---

Now substitute the values of $\sin A$, $\cos B$, $\cos A$, and $\sin B$ into the formula for $\sin(A+B)$:

$\sin(A+B) = \left(\frac{3}{5}\right) \times \left(\frac{5}{13}\right) + \left(\frac{4}{5}\right) \times \left(\frac{12}{13}\right)$

$\sin(A+B) = \frac{3 \times 5}{5 \times 13} + \frac{4 \times 12}{5 \times 13}$

$\sin(A+B) = \frac{15}{65} + \frac{48}{65}$

Add the fractions with the common denominator:

$\sin(A+B) = \frac{15 + 48}{65}$

$\sin(A+B) = \frac{63}{65}$

---

Comparing our result with the given options, we find that the value of $\sin(A+B)$ is $\frac{63}{65}$, which corresponds to option (B).

---

The final answer is (B) $\frac{63}{65}$.

We need to find which of the given angles is coterminal with $-45^\circ$.

---

Coterminal angles are angles in standard position that have the same terminal side. They differ by an integer multiple of a full circle, which is $360^\circ$ or $2\pi$ radians.

To find a coterminal angle for a given angle $\theta$, we can add or subtract integer multiples of $360^\circ$.

---

The given angle is $\theta = -45^\circ$. We need to check if any of the options can be obtained by adding $n \times 360^\circ$ to $-45^\circ$ for some integer $n$.

---

Let's check option (B): $315^\circ$.

Is $315^\circ = -45^\circ + n \times 360^\circ$ for some integer $n$?

Add $45^\circ$ to both sides:

$315^\circ + 45^\circ = n \times 360^\circ$

$360^\circ = n \times 360^\circ$

Dividing both sides by $360^\circ$, we get $n = 1$.

Since $n=1$ is an integer, $315^\circ$ is coterminal with $-45^\circ$.

---

Let's briefly check the other options for verification:

• Option (A) $45^\circ$: $45^\circ = -45^\circ + n \times 360^\circ \implies 90^\circ = n \times 360^\circ \implies n = \frac{90}{360} = \frac{1}{4}$, which is not an integer.
• Option (C) $-135^\circ$: $-135^\circ = -45^\circ + n \times 360^\circ \implies -90^\circ = n \times 360^\circ \implies n = \frac{-90}{360} = -\frac{1}{4}$, which is not an integer.
• Option (D) $225^\circ$: $225^\circ = -45^\circ + n \times 360^\circ \implies 270^\circ = n \times 360^\circ \implies n = \frac{270}{360} = \frac{3}{4}$, which is not an integer.

---

Only $315^\circ$ is coterminal with $-45^\circ$ among the given options.

---

The final answer is (B) $315^\circ$.

Given:

Radius of the circle, $r = 5$ cm.

Central angle subtended by the arc, $\theta = 30^\circ$.

---

To Find:

The length of the arc, $l$.

---

Solution:

The formula for the length of an arc ($l$) of a circle with radius $r$ subtending a central angle $\theta$ (in radians) is:

Important Note: The angle $\theta$ in this formula must be in radians, not degrees.

---

The given angle is $\theta = 30^\circ$. We need to convert this angle from degrees to radians.

The conversion formula from degrees to radians is: $\text{Radians} = \text{Degrees} \times \frac{\pi}{180^\circ}$.

Convert $30^\circ$ to radians:

Simplify the expression:

---

Now substitute the values of $r$ and $\theta$ (in radians) into the arc length formula (i):

$r = 5$ cm

$\theta = \frac{\pi}{6}$ radians

Multiply the values:

---

The length of the arc is $\frac{5\pi}{6}$ cm.

---

Comparing our result with the given options, we find that the length of the arc is $\frac{5\pi}{6}$ cm, which corresponds to option (B).

---

The final answer is (B) $\frac{5\pi}{6}$ cm.

Given:

The function is $f(x) = \frac{1}{\sin x}$.

---

To Find:

The domain of the function $f(x)$.

---

Solution:

The domain of a function is the set of all possible input values (values of $x$) for which the function is defined.

The given function is a rational function where the numerator is 1 and the denominator is $\sin x$.

A rational function is undefined when its denominator is equal to zero.

Therefore, the function $f(x) = \frac{1}{\sin x}$ is defined for all real values of $x$ where the denominator, $\sin x$, is not equal to zero.

We need to find the values of $x$ for which $\sin x = 0$.

The general solution for the trigonometric equation $\sin x = 0$ is given by:

---

These are the values of $x$ that must be excluded from the domain of $f(x)$.

The domain of $f(x)$ is the set of all real numbers $\mathbb{R}$, excluding the set of values $\{n\pi : n \in \mathbb{Z}\}$.

Domain of $f(x) = \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$.

---

Comparing our result with the given options, we find that the domain matches option (A).

• (A) $\mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$: Excludes values where $\sin x = 0$. Correct.
• (B) $\mathbb{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$: Excludes odd multiples of $\frac{\pi}{2}$, which are the values where $\cos x = 0$ (and $\sin x = \pm 1$). Incorrect.
• (C) $\mathbb{R} - \{n\frac{\pi}{2} : n \in \mathbb{Z}\}$: Excludes all multiples of $\frac{\pi}{2}$, both even ($n\pi$) and odd ($(2n+1)\frac{\pi}{2}$). This excludes points where $\cos x = 0$ as well. Incorrect.
• (D) $\mathbb{R}$: States that the function is defined for all real numbers. Incorrect.

---

The final answer is (A) $\mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$.

Given:

The expression is $\sin 15^\circ \cos 75^\circ$.

---

To Find:

The value of the given expression.

---

Solution:

We can use the complementary angle identity: $\cos(90^\circ - \theta) = \sin \theta$.

Let $\theta = 15^\circ$. Then $90^\circ - 15^\circ = 75^\circ$.

So, $\cos 75^\circ = \cos(90^\circ - 15^\circ) = \sin 15^\circ$.

---

Substitute this into the given expression:

$\sin 15^\circ \cos 75^\circ = \sin 15^\circ \times (\sin 15^\circ)$

---

To find the value of $\sin^2 15^\circ$, we can first find $\sin 15^\circ$. We know $\sin 15^\circ = \sin(45^\circ - 30^\circ)$.

Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:

$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$

$\sin 15^\circ = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right)$

$\sin 15^\circ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

---

Now square this value to find $\sin^2 15^\circ$ (from equation (i)):

$\sin^2 15^\circ = \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2$

$\sin^2 15^\circ = \frac{(\sqrt{6})^2 - 2(\sqrt{6})(\sqrt{2}) + (\sqrt{2})^2}{4^2}$

$\sin^2 15^\circ = \frac{6 - 2\sqrt{12} + 2}{16}$

$\sin^2 15^\circ = \frac{8 - 2(2\sqrt{3})}{16}$

$\sin^2 15^\circ = \frac{8 - 4\sqrt{3}}{16}$

Factor out 4 from the numerator:

Cancel the common factor 4:

---

The value of $\sin 15^\circ \cos 75^\circ$ is $\frac{2 - \sqrt{3}}{4}$.

Let's compare this result with the given options:

• (A) $\frac{1}{4}$
• (B) $\frac{1}{2} = \frac{2}{4}$
• (C) $\frac{\sqrt{3}}{4}$
• (D) $\frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{4}$

The calculated value $\frac{2 - \sqrt{3}}{4}$ does not match any of the given options.

---

Assumption based on options:

Given that this is a multiple-choice question and the calculated value does not match any option, it is highly probable there is a typo in the question or the options.

Let us consider if the question was intended to be $\sin 15^\circ \sin 75^\circ$.

Using the complementary angle identity, $\sin 75^\circ = \cos(90^\circ - 75^\circ) = \cos 15^\circ$.

So, the expression becomes $\sin 15^\circ \sin 75^\circ = \sin 15^\circ \cos 15^\circ$.

We can use the double angle identity: $\sin(2\theta) = 2 \sin \theta \cos \theta$, which implies $\sin \theta \cos \theta = \frac{1}{2} \sin(2\theta)$.

Let $\theta = 15^\circ$. Then $2\theta = 2 \times 15^\circ = 30^\circ$.

$\sin 15^\circ \cos 15^\circ = \frac{1}{2} \sin(2 \times 15^\circ) = \frac{1}{2} \sin 30^\circ$.

We know the value of $\sin 30^\circ = \frac{1}{2}$.

$\sin 15^\circ \cos 15^\circ = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

---

This result, $\frac{1}{4}$, matches option (A).

Based on the likely intended question, the answer is $\frac{1}{4}$.

---

Concluding based on the assumption that the intended question leads to one of the given options, the most plausible intended question is $\sin 15^\circ \sin 75^\circ$ (or equivalent), which results in option (A).

---

The final answer, assuming a likely typo in the question leading to a valid option, is (A) $\frac{1}{4}$.

Given:

The trigonometric equation is $\cos x = -1$.

---

To Find:

The general solution for $x$, where $n$ is an integer.

---

Solution:

We need to find the values of $x$ for which $\cos x = -1$.

We know that the principal value of $x$ in the interval $[0, 2\pi)$ for which $\cos x = -1$ is $x = \pi$.

---

The general solution for the equation $\cos x = \cos \alpha$ is given by:

---

In our case, $\cos x = -1 = \cos(\pi)$. So, $\alpha = \pi$.

Substitute $\alpha = \pi$ into the general solution formula:

$x = 2n\pi \pm \pi$

---

We can factor out $\pi$ from the right side:

$x = (2n \pm 1)\pi$

---

This expression represents angles that are odd multiples of $\pi$ (e.g., when $n=0$, $x = \pm \pi$; when $n=1$, $x = 2\pi \pm \pi = 3\pi, \pi$; when $n=-1$, $x = -2\pi \pm \pi = -3\pi, -\pi$). Any odd integer can be written in the form $2n+1$ or $2n-1$. Therefore, the solution can be written as $x = (2n+1)\pi$ or $x = (2n-1)\pi$. Both forms represent the same set of odd multiples of $\pi$. The form $(2n+1)\pi$ covers all odd integers for varying $n \in \mathbb{Z}$.

Thus, the general solution is $x = (2n+1)\pi$, where $n \in \mathbb{Z}$.

---

Comparing this general solution with the given options:

• (A) $x = n\pi$: This represents all integer multiples of $\pi$, including even multiples (where $\cos x = 1$) and odd multiples (where $\cos x = -1$). This is incorrect as we only want $\cos x = -1$.
• (B) $x = 2n\pi$: This represents even multiples of $\pi$, where $\cos x = \cos(2n\pi) = 1$. This is incorrect.
• (C) $x = (2n+1)\pi$: This represents odd multiples of $\pi$, where $\cos x = \cos((2n+1)\pi) = -1$. This is correct.
• (D) $x = (n + \frac{1}{2})\pi = \frac{(2n+1)\pi}{2}$: This represents odd multiples of $\frac{\pi}{2}$, where $\cos x = 0$. This is incorrect.

---

The general solution of $\cos x = -1$ is $x = (2n+1)\pi$, where $n \in \mathbb{Z}$.

---

The final answer is (C) $x = (2n+1)\pi, n \in \mathbb{Z}$.

Given:

The function is $f(x) = \tan(\frac{x}{2})$.

---

To Find:

The period of the function $f(x)$.

---

Solution:

The general form of a tangent function is $g(x) = A \tan(Bx + C) + D$.

The period of such a function is given by the formula:

Note that the basic period of the tangent function $\tan x$ is $\pi$, unlike sine and cosine functions which have a basic period of $2\pi$.

---

Our given function is $f(x) = \tan(\frac{x}{2})$.

We can rewrite this as $f(x) = \tan(\frac{1}{2}x)$.

Comparing this with the form $\tan(Bx)$, we can identify the value of $B$.

---

Now, we substitute the value of $B$ into the period formula (i):

Period $= \frac{\pi}{|1/2|}$

Period $= \frac{\pi}{1/2}$

To divide by a fraction, we multiply by its reciprocal:

Period $= \pi \times 2$

---

Thus, the period of the function $f(x) = \tan(\frac{x}{2})$ is $2\pi$.

---

Comparing our result with the given options, we find that the period is $2\pi$, which corresponds to option (C).

---

The final answer is (C) $2\pi$.

Given:

We are given that $\sin x = \frac{3}{5}$ and $\cos y = \frac{4}{5}$.

Angles $x$ and $y$ are in the first quadrant ($x, y \in (0, \frac{\pi}{2})$).

---

To Find:

The value of $\sin(x-y)$.

---

Solution:

We use the trigonometric identity for the sine of the difference of two angles:

$\sin(A-B) = \sin A \cos B - \cos A \sin B$

For our problem, $A=x$ and $B=y$. So, we need to find $\sin x, \cos y, \cos x, \sin y$.

We are given $\sin x = \frac{3}{5}$ and $\cos y = \frac{4}{5}$.

---

Since $x$ is in the first quadrant, $\cos x$ is positive.

Using the identity $\sin^2 x + \cos^2 x = 1$:

$\cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25-9}{25} = \frac{16}{25}$

$\cos x = \sqrt{\frac{16}{25}}$

---

Since $y$ is in the first quadrant, $\sin y$ is positive.

Using the identity $\sin^2 y + \cos^2 y = 1$:

$\sin^2 y = 1 - \cos^2 y = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{25-16}{25} = \frac{9}{25}$

$\sin y = \sqrt{\frac{9}{25}}$

---

Now substitute the values of $\sin x, \cos y, \cos x, \sin y$ into the formula for $\sin(x-y)$:

$\sin(x-y) = \sin x \cos y - \cos x \sin y$

$\sin(x-y) = \left(\frac{3}{5}\right) \times \left(\frac{4}{5}\right) - \left(\frac{4}{5}\right) \times \left(\frac{3}{5}\right)$

$\sin(x-y) = \frac{3 \times 4}{5 \times 5} - \frac{4 \times 3}{5 \times 5}$

$\sin(x-y) = \frac{12}{25} - \frac{12}{25}$

$\sin(x-y) = 0$

---

The value of $\sin(x-y)$ is 0.

---

Comparing our result with the given options, we find that the value is 0, which corresponds to option (A).

---

Note:

We have $\sin x = \frac{3}{5}$ and $\cos x = \frac{4}{5}$. For an acute angle $x$, this means $x = \arcsin(\frac{3}{5})$ and $x = \arccos(\frac{4}{5})$.

We have $\cos y = \frac{4}{5}$ and $\sin y = \frac{3}{5}$. For an acute angle $y$, this means $y = \arccos(\frac{4}{5})$ and $y = \arcsin(\frac{3}{5})$.

Since $x$ and $y$ are acute angles, and $\sin x = \sin y$ and $\cos x = \cos y$, it implies that $x = y$.

If $x = y$, then $\sin(x-y) = \sin(x-x) = \sin(0) = 0$. This confirms our calculated result.

---

The final answer is (A) 0.

Given:

The trigonometric equation is $\tan x = 0$.

---

To Find:

The value among the options that is NOT a solution to $\tan x = 0$.

---

Solution:

The tangent function is defined as $\tan x = \frac{\sin x}{\cos x}$.

For $\tan x$ to be equal to 0, the numerator $\sin x$ must be equal to 0, and the denominator $\cos x$ must not be equal to 0 (otherwise, the expression would be undefined, not 0).

---

So, we are looking for values of $x$ such that:

AND

---

The general solution for the equation $\sin x = 0$ is:

---

Now let's check if these values satisfy the condition $\cos x \neq 0$.

For $x = n\pi$, the value of $\cos x$ is $\cos(n\pi)$.

If $n$ is an even integer (e.g., $n=0, 2, -2$), $x = 2k\pi$ for some integer $k$, and $\cos(2k\pi) = 1 \neq 0$.

If $n$ is an odd integer (e.g., $n=1, -1, 3$), $x = (2k+1)\pi$ for some integer $k$, and $\cos((2k+1)\pi) = -1 \neq 0$.

So, for all values $x = n\pi$ where $n \in \mathbb{Z}$, $\cos x$ is either 1 or -1, which is never 0. Thus, condition (ii) is satisfied when condition (i) is satisfied.

Therefore, the solutions to $\tan x = 0$ are exactly the values $x = n\pi$, where $n$ is any integer.

---

Now let's check the given options to see which one is NOT of the form $n\pi$:

• (A) $x = 0$. This is of the form $n\pi$ with $n=0$ ($0 = 0 \times \pi$). So, $\tan(0) = 0$. This is a solution.
• (B) $x = \pi$. This is of the form $n\pi$ with $n=1$ ($\pi = 1 \times \pi$). So, $\tan(\pi) = 0$. This is a solution.
• (C) $x = 2\pi$. This is of the form $n\pi$ with $n=2$ ($2\pi = 2 \times \pi$). So, $\tan(2\pi) = 0$. This is a solution.
• (D) $x = \pi/2$. This is not of the form $n\pi$ for any integer $n$. At $x = \pi/2$, $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$. Thus, $\tan(\pi/2) = \frac{1}{0}$, which is undefined. An undefined value is not equal to 0. Therefore, $x = \pi/2$ is NOT a solution to $\tan x = 0$.

---

The value among the options that is not a solution to $\tan x = 0$ is $\pi/2$.

---

The final answer is (D) $\pi/2$.

We are given an Assertion (A) and a Reason (R) and need to determine their truthfulness and if the Reason correctly explains the Assertion.

---

Assertion (A): The general solution of $\cos x = 1$ is $x = 2n\pi$, where $n \in \mathbb{Z}$.

To find the general solution of $\cos x = 1$, we first find the principal value of $x$ for which $\cos x = 1$. This occurs at $x=0$ (or $x=2\pi$, etc.).

The general solution for the equation $\cos x = \cos \alpha$ is given by $x = 2n\pi \pm \alpha$, where $n \in \mathbb{Z}$.

In this case, $\cos x = 1 = \cos(0)$. So, we can take $\alpha = 0$.

This means that the values of $x$ for which $\cos x = 1$ are the even multiples of $\pi$ (e.g., $0, 2\pi, -2\pi, 4\pi$, etc.).

So, Assertion (A) is True.

---

Reason (R): The cosine function is 1 at even multiples of $\pi$.

Let's evaluate the cosine function at even multiples of $\pi$. Even multiples of $\pi$ are represented by $2n\pi$, where $n \in \mathbb{Z}$.

• For $n=0$, $2n\pi = 0$, $\cos(0) = 1$.
• For $n=1$, $2n\pi = 2\pi$, $\cos(2\pi) = 1$.
• For $n=2$, $2n\pi = 4\pi$, $\cos(4\pi) = 1$.
• For $n=-1$, $2n\pi = -2\pi$, $\cos(-2\pi) = 1$.

In general, for any integer $n$, $\cos(2n\pi) = 1$.

So, Reason (R) is True.

---

Now, we need to check if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that the set of all solutions to $\cos x = 1$ is precisely the set of even multiples of $\pi$. Reason (R) states that the cosine function has the value 1 at exactly these even multiples of $\pi$. This is the fundamental property that leads to the general solution $x = 2n\pi$ for $\cos x = 1$. Therefore, Reason (R) correctly explains Assertion (A).

---

Since both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A), the correct option is (A).

---

The final answer is (A) Both A and R are true and R is the correct explanation of A.

Given:

We need to find the value of $\text{cosec}(\pi/6)$.

---

To Find:

The value of the expression $\text{cosec}(\pi/6)$.

---

Solution:

The cosecant function is the reciprocal of the sine function.

---

In this problem, $x = \pi/6$.

The angle $\pi/6$ radians is equivalent to $30^\circ$ ($1 \text{ radian} = \frac{180^\circ}{\pi}$).

$\pi/6 \text{ radians} = \frac{\pi}{6} \times \frac{180^\circ}{\pi} = \frac{180^\circ}{6} = 30^\circ$.

---

We need the value of $\sin(\pi/6)$ or $\sin(30^\circ)$.

So, $\sin(\pi/6) = \frac{1}{2}$.

---

Now, substitute this value into the definition of cosecant (i):

$\text{cosec}(\pi/6) = \frac{1}{\sin(\pi/6)} = \frac{1}{1/2}$

To divide by a fraction, we multiply by its reciprocal:

$\text{cosec}(\pi/6) = 1 \times \frac{2}{1} = 2$

---

The value of $\text{cosec}(\pi/6)$ is 2.

---

Comparing our result with the given options, we find that the value is 2, which corresponds to option (C).

---

The final answer is (C) 2.

We need to identify which of the given trigonometric identities is incorrect.

---

Let's examine each option:

(A) $\sin 2x = 2 \sin x \cos x$

This is a fundamental double angle identity for the sine function. This identity is Correct.

---

(B) $\cos 2x = 2\cos^2 x - 1$

This is one of the standard double angle identities for the cosine function, derived from the identity $\cos 2x = \cos^2 x - \sin^2 x$ by substituting $\sin^2 x = 1 - \cos^2 x$. This identity is Correct.

---

(C) $\tan 2x = \frac{2 \tan x}{1 + \tan^2 x}$

Let's compare this to the known double angle identity for tangent. The correct identity is $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$.

Also, let's consider the expression on the right side: $\frac{2 \tan x}{1 + \tan^2 x}$. We know that $1 + \tan^2 x = \sec^2 x$.

So, $\frac{2 \tan x}{1 + \tan^2 x} = \frac{2 (\frac{\sin x}{\cos x})}{\sec^2 x} = \frac{\frac{2 \sin x}{\cos x}}{\frac{1}{\cos^2 x}} = \frac{2 \sin x}{\cos x} \times \cos^2 x = 2 \sin x \cos x$.

This is equal to $\sin 2x$ (from option A).

Therefore, option (C) states that $\tan 2x = \sin 2x$, which is generally false. This identity is Incorrect.

---

(D) $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$

From our evaluation of option (C), we found that the right side $\frac{2 \tan x}{1 + \tan^2 x}$ simplifies to $2 \sin x \cos x$, which is equal to $\sin 2x$ (as seen in option A). This identity is Correct.

---

Based on the analysis, the incorrect identity among the given options is (C).

---

The final answer is (C) $\tan 2x = \frac{2 \tan x}{1 + \tan^2 x}$.

We need to identify which of the given angles lie in the third quadrant.

---

The quadrants in the standard Cartesian coordinate system are defined based on the angle measured counterclockwise from the positive x-axis.

• Quadrant I: $(0^\circ, 90^\circ)$ or $(0, \pi/2)$ radians
• Quadrant II: $(90^\circ, 180^\circ)$ or $(\pi/2, \pi)$ radians
• Quadrant III: $(180^\circ, 270^\circ)$ or $(\pi, 3\pi/2)$ radians
• Quadrant IV: $(270^\circ, 360^\circ)$ or $(3\pi/2, 2\pi)$ radians

Angles can also be measured clockwise, resulting in negative values. A clockwise angle $\theta$ is coterminal with a counterclockwise angle $360^\circ + \theta$ (in degrees) or $2\pi + \theta$ (in radians).

---

Let's examine each option:

(A) $210^\circ$

We check if $210^\circ$ is within the range of the third quadrant $(180^\circ, 270^\circ)$.

The inequality is true.

So, $210^\circ$ lies in the third quadrant.

---

(B) $-140^\circ$

This is a negative angle. We can add $360^\circ$ to find a coterminal positive angle:

$-140^\circ + 360^\circ = 220^\circ$

Now we check if $220^\circ$ is within the range of the third quadrant $(180^\circ, 270^\circ)$.

The inequality is true.

Alternatively, the third quadrant for negative angles is $(-180^\circ, -90^\circ)$.

The inequality is true.

So, $-140^\circ$ lies in the third quadrant.

---

(C) $5\pi/4$ radians

We check if $5\pi/4$ is within the range of the third quadrant $(\pi, 3\pi/2)$ radians.

Convert $\pi$ and $3\pi/2$ to fractions with denominator 4:

$\pi = \frac{4\pi}{4}$

$3\pi/2 = \frac{3\pi \times 2}{2 \times 2} = \frac{6\pi}{4}$

Now we check the inequality:

The inequality is true.

So, $5\pi/4$ radians lies in the third quadrant.

---

(D) $7\pi/6$ radians

We check if $7\pi/6$ is within the range of the third quadrant $(\pi, 3\pi/2)$ radians.

Convert $\pi$ and $3\pi/2$ to fractions with denominator 6:

$\pi = \frac{6\pi}{6}$

$3\pi/2 = \frac{3\pi \times 3}{2 \times 3} = \frac{9\pi}{6}$

Now we check the inequality:

The inequality is true.

So, $7\pi/6$ radians lies in the third quadrant.

---

All four given angles ($210^\circ$, $-140^\circ$, $5\pi/4$ radians, and $7\pi/6$ radians) lie in the third quadrant.

Since this is a Multiple Correct Answer(s) question, all options that are correct should be selected.

---

The correct options are (A) $210^\circ$, (B) $-140^\circ$, (C) $5\pi/4$ radians, and (D) $7\pi/6$ radians.

Given:

The expression is $\sin(-\frac{11\pi}{3})$.

---

To Find:

The value of the given expression.

---

Solution:

We need to evaluate $\sin(-\frac{11\pi}{3})$.

First, we can use the property that the sine function is an odd function, which means $\sin(-\theta) = -\sin \theta$.

---

Next, we can simplify the angle $\frac{11\pi}{3}$ by expressing it as a sum or difference involving a multiple of $2\pi$ (the period of the sine function) and a standard angle.

We can write $\frac{11\pi}{3}$ as a mixed number or find the closest multiple of $2\pi$.

$2\pi = \frac{6\pi}{3}$, $4\pi = \frac{12\pi}{3}$.

$\frac{11\pi}{3}$ is close to $\frac{12\pi}{3} = 4\pi$.

We can write $\frac{11\pi}{3} = 4\pi - \frac{\pi}{3}$.

---

Now substitute this into the expression:

$-\sin(\frac{11\pi}{3}) = -\sin(4\pi - \frac{\pi}{3})$

---

We can use the periodicity of the sine function: $\sin(2n\pi - \theta) = \sin(-\theta) = -\sin \theta$ for any integer $n$. Here, $4\pi = 2(2)\pi$, so $n=2$.

---

Substitute this back into the main expression:

$-\sin(4\pi - \frac{\pi}{3}) = -(-\sin(\frac{\pi}{3}))$

$-(-\sin(\frac{\pi}{3})) = \sin(\frac{\pi}{3})$

---

Now we evaluate the value of $\sin(\frac{\pi}{3})$.

The angle $\frac{\pi}{3}$ radians is equal to $60^\circ$.

---

Thus, the value of $\sin(-\frac{11\pi}{3})$ is $\frac{\sqrt{3}}{2}$.

---

Comparing our result with the given options, we find that the value is $\frac{\sqrt{3}}{2}$, which corresponds to option (A).

---

The final answer is (A) $\frac{\sqrt{3}}{2}$.

To convert an angle from degrees and minutes to radians, we first convert the entire angle into degrees and then use the conversion formula from degrees to radians.

---

We know that $1^\circ = 60$ minutes ($60'$).

Therefore, $1' = \left(\frac{1}{60}\right)^\circ$.

The given angle is $40^\circ 20'$.

---

First, convert the minutes part into degrees:

$20' = \left(20 \times \frac{1}{60}\right)^\circ = \left(\frac{20}{60}\right)^\circ = \left(\frac{1}{3}\right)^\circ$.

---

Now, add this to the degree part:

$40^\circ 20' = \left(40 + \frac{1}{3}\right)^\circ$.

Combine the terms within the parentheses:

$40 + \frac{1}{3} = \frac{40 \times 3 + 1}{3} = \frac{120 + 1}{3} = \frac{121}{3}$.

So, $40^\circ 20' = \left(\frac{121}{3}\right)^\circ$.

---

Now, we convert this angle from degrees to radians.

The conversion formula is $1^\circ = \frac{\pi}{180}$ radians.

Therefore, $\left(\frac{121}{3}\right)^\circ = \left(\frac{121}{3} \times \frac{\pi}{180}\right)$ radians.

---

Multiply the terms:

$\frac{121}{3} \times \frac{\pi}{180} = \frac{121 \times \pi}{3 \times 180} = \frac{121\pi}{540}$.

---

Thus, $40^\circ 20'$ in radian measure is $\frac{121\pi}{540}$ radians.

The final answer is $\boxed{\frac{121\pi}{540} \text{ radians}}$.

Given:

Radius of the circle, $r = 100 \text{ cm}$.

Length of the arc, $l = 22 \text{ cm}$.

Value of $\pi = \frac{22}{7}$.

---

To Find:

The degree measure of the angle subtended at the centre.

---

Solution:

The relation between the arc length ($l$), radius ($r$), and the angle ($\theta$) subtended at the centre (in radians) is given by the formula:

$l = r\theta$

---

We need to find the angle $\theta$. Rearranging the formula, we get:

$\theta = \frac{l}{r}$

---

Substitute the given values of $l$ and $r$ into the formula:

$\theta = \frac{22 \text{ cm}}{100 \text{ cm}}$

$\theta = \frac{22}{100}$ radians

---

Now, we need to convert this angle from radians to degrees.

We know that $\pi$ radians is equal to $180^\circ$.

So, $1$ radian $= \left(\frac{180}{\pi}\right)^\circ$.

---

To convert $\frac{22}{100}$ radians to degrees, we multiply it by the conversion factor $\frac{180}{\pi}$:

Angle in degrees $= \left(\frac{22}{100} \times \frac{180}{\pi}\right)^\circ$

---

Substitute the given value of $\pi = \frac{22}{7}$ into the expression:

Angle in degrees $= \left(\frac{22}{100} \times \frac{180}{\frac{22}{7}}\right)^\circ$

Angle in degrees $= \left(\frac{22}{100} \times 180 \times \frac{7}{22}\right)^\circ$

---

Now, we simplify the expression. We can cancel out the common factor of 22 in the numerator and the denominator:

Angle in degrees $= \left(\frac{\cancel{22}}{100} \times 180 \times \frac{7}{\cancel{22}}\right)^\circ$

Angle in degrees $= \left(\frac{1}{100} \times 180 \times 7\right)^\circ$

---

Multiply the numbers in the numerator:

Angle in degrees $= \left(\frac{180 \times 7}{100}\right)^\circ$

Angle in degrees $= \left(\frac{1260}{100}\right)^\circ$

---

Divide the numbers:

Angle in degrees $= \left(\frac{126}{10}\right)^\circ$

Angle in degrees $= 12.6^\circ$

---

The angle is $12.6^\circ$. To express this in degrees and minutes, we separate the decimal part.

$12.6^\circ = 12^\circ + 0.6^\circ$

We convert the decimal part from degrees to minutes. We know that $1^\circ = 60'$.

$0.6^\circ = 0.6 \times 60'$

$0.6 \times 60 = 36$

So, $0.6^\circ = 36'$.

---

Therefore, $12.6^\circ = 12^\circ 36'$.

The degree measure of the angle subtended at the centre is $12^\circ 36'$.

The final answer is $\boxed{12^\circ 36'}$.

Given:

$\sin x = \frac{3}{5}$

The angle $x$ lies in the second quadrant.

---

To Find:

The values of the other five trigonometric functions: $\cos x, \tan x, \text{cosec } x, \sec x, \cot x$.

---

Solution:

We are given $\sin x = \frac{3}{5}$. Since $x$ is in the second quadrant, $\sin x$ is positive, which is consistent with the given value.

---

We use the fundamental trigonometric identity:

$\sin^2 x + \cos^2 x = 1$

Simplify the square of $\sin x$:

$\frac{9}{25} + \cos^2 x = 1$

---

Now, isolate $\cos^2 x$:

$\cos^2 x = 1 - \frac{9}{25}$

$\cos^2 x = \frac{25}{25} - \frac{9}{25}$

$\cos^2 x = \frac{25 - 9}{25}$

$\cos^2 x = \frac{16}{25}$

---

Taking the square root of both sides:

$\cos x = \pm \sqrt{\frac{16}{25}}$

$\cos x = \pm \frac{4}{5}$

---

Since $x$ lies in the second quadrant, the value of $\cos x$ is negative.

---

Next, we find $\tan x$ using the identity $\tan x = \frac{\sin x}{\cos x}$:

$\tan x = \frac{\frac{3}{5}}{-\frac{4}{5}}$

$\tan x = \frac{3}{5} \times \left(-\frac{5}{4}\right)$

$\tan x = -\frac{3 \times 5}{5 \times 4}$

$\tan x = -\frac{\cancel{3 \times 5}^{\;3}}{\cancel{5 \times 4}_{\;4}}$

$\tan x = -\frac{3}{4}$

---

Now, we find the reciprocal trigonometric functions:

$\text{cosec } x = \frac{1}{\sin x}$

$\text{cosec } x = \frac{1}{\frac{3}{5}} = \frac{5}{3}$

---

$\sec x = \frac{1}{\cos x}$

$\sec x = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

---

$\cot x = \frac{1}{\tan x}$

$\cot x = \frac{1}{-\frac{3}{4}} = -\frac{4}{3}$

---

Summary of the values of the other five trigonometric functions:

$\cos x = -\frac{4}{5}$

$\tan x = -\frac{3}{4}$

$\text{cosec } x = \frac{5}{3}$

$\sec x = -\frac{5}{4}$

$\cot x = -\frac{4}{3}$

To Prove:

$\sin(x+y) = \sin x \cos y + \cos x \sin y$

---

Given Identity:

$\cos(x+y) = \cos x \cos y - \sin x \sin y$

---

Proof:

We know the co-function identity that states $\sin \theta = \cos\left(\frac{\pi}{2} - \theta\right)$.

Let $\theta = x+y$. Using the identity, we can write:

$\sin(x+y) = \cos\left(\frac{\pi}{2} - (x+y)\right)$

---

Rearrange the argument of the cosine function:

$\sin(x+y) = \cos\left(\left(\frac{\pi}{2} - x\right) - y\right)$

---

Now, we use the identity for the cosine of a difference, $\cos(A-B) = \cos A \cos B + \sin A \sin B$. This identity can be derived from the given identity $\cos(x+y) = \cos x \cos y - \sin x \sin y$ by replacing $y$ with $-y$.

Applying $\cos(A-B) = \cos A \cos B + \sin A \sin B$ with $A = \frac{\pi}{2} - x$ and $B = y$, we get:

$\sin(x+y) = \cos\left(\frac{\pi}{2} - x\right) \cos y + \sin\left(\frac{\pi}{2} - x\right) \sin y$

---

We use the co-function identities again:

$\cos\left(\frac{\pi}{2} - x\right) = \sin x$

$\sin\left(\frac{\pi}{2} - x\right) = \cos x$

---

Substitute these results back into the expression for $\sin(x+y)$:

$\sin(x+y) = (\sin x) \cos y + (\cos x) \sin y$

$\sin(x+y) = \sin x \cos y + \cos x \sin y$

---

This proves the identity $\sin(x+y) = \sin x \cos y + \cos x \sin y$ using the given identity for $\cos(x+y)$.

Given Equation:

$\sin x = \frac{\sqrt{3}}{2}$

---

To Find:

The principal solutions of the equation.

---

Solution:

The principal solutions of a trigonometric equation are the solutions that lie in the interval $[0, 2\pi)$.

---

We are given $\sin x = \frac{\sqrt{3}}{2}$.

We know that $\sin \theta = \frac{\sqrt{3}}{2}$ for $\theta = \frac{\pi}{3}$.

So, one possible value for $x$ is $\frac{\pi}{3}$.

---

The sine function is positive in the first quadrant and the second quadrant.

---

In the first quadrant, the principal solution is the reference angle itself.

$x = \frac{\pi}{3}$

---

In the second quadrant, the angle with the same sine value is given by $\pi - \theta$, where $\theta$ is the reference angle $\frac{\pi}{3}$.

$x = \pi - \frac{\pi}{3}$

To simplify, find a common denominator:

$x = \frac{3\pi}{3} - \frac{\pi}{3} = \frac{3\pi - \pi}{3} = \frac{2\pi}{3}$

---

Both $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ lie within the principal range $[0, 2\pi)$.

---

Thus, the principal solutions of the equation $\sin x = \frac{\sqrt{3}}{2}$ are $x = \frac{\pi}{3}$ and $x = \frac{2\pi}{3}$.

The final answer is $\boxed{x = \frac{\pi}{3}, \frac{2\pi}{3}}$.

Given:

Angle in radian measure = 6 radians.

Value of $\pi = \frac{22}{7}$.

---

To Find:

The degree measure of the angle.

---

Solution:

The conversion formula from radians to degrees is:

Degrees = Radians $\times \frac{180^\circ}{\pi}$

---

Substitute the given value of the angle in radians (6) into the formula:

Degrees $= 6 \times \frac{180^\circ}{\pi}$

---

Substitute the given value of $\pi = \frac{22}{7}$ into the expression:

Degrees $= 6 \times \frac{180^\circ}{\frac{22}{7}}$

Degrees $= 6 \times 180^\circ \times \frac{7}{22}$

---

Now, we simplify the expression. We can cancel out a factor of 2 from 6 and 22:

Degrees $= \cancel{6}^{\;3} \times 180^\circ \times \frac{7}{\cancel{22}_{\;11}}$

Degrees $= 3 \times 180^\circ \times \frac{7}{11}$

---

Multiply the numbers in the numerator:

Degrees $= \frac{3 \times 180 \times 7}{11}^\circ$

Degrees $= \frac{540 \times 7}{11}^\circ$

Degrees $= \frac{3780}{11}^\circ$

---

To express this in degrees, minutes, and seconds, we perform division:

Divide 3780 by 11:

$3780 \div 11 = 343$ with a remainder of 7.

So, $\frac{3780}{11}^\circ = 343^\circ + \frac{7}{11}^\circ$.

---

Convert the fractional part of degrees to minutes. We know $1^\circ = 60'$.

$\frac{7}{11}^\circ = \left(\frac{7}{11} \times 60\right)' = \frac{420}{11}'$.

---

Divide 420 by 11 to convert minutes into whole minutes and a fractional part:

$420 \div 11 = 38$ with a remainder of 2.

So, $\frac{420}{11}' = 38' + \frac{2}{11}'$.

---

Convert the fractional part of minutes to seconds. We know $1' = 60''$.

$\frac{2}{11}' = \left(\frac{2}{11} \times 60\right)'' = \frac{120}{11}''$.

---

Divide 120 by 11 to find seconds:

$120 \div 11 \approx 10.91''$. We can approximate this to 11 seconds.

So, $\frac{120}{11}'' \approx 11''$.

---

Combining the degrees, minutes, and seconds, we get:

6 radians $\approx 343^\circ 38' 11''$.

The degree measure of 6 radians is approximately $343^\circ 38' 11''$.

To Find:

The value of $\cos 15^\circ$.

---

Solution:

We can express $15^\circ$ as the difference of two standard angles whose trigonometric values are known. For example, $15^\circ = 45^\circ - 30^\circ$ or $15^\circ = 60^\circ - 45^\circ$.

Let's use the angle difference formula for cosine:

$\cos(A - B) = \cos A \cos B + \sin A \sin B$

---

Let $A = 45^\circ$ and $B = 30^\circ$. Then $A - B = 45^\circ - 30^\circ = 15^\circ$.

We need the trigonometric values for $45^\circ$ and $30^\circ$:

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

$\sin 30^\circ = \frac{1}{2}$

---

Substitute these values into the formula for $\cos(A - B)$:

$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$

---

Multiply the terms:

$\cos 15^\circ = \frac{1 \times \sqrt{3}}{\sqrt{2} \times 2} + \frac{1 \times 1}{\sqrt{2} \times 2}$

$\cos 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$

---

Combine the fractions since they have a common denominator:

$\cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

---

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$\cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\cos 15^\circ = \frac{(\sqrt{3} + 1)\sqrt{2}}{2 \times (\sqrt{2})^2}$

$\cos 15^\circ = \frac{\sqrt{3} \times \sqrt{2} + 1 \times \sqrt{2}}{2 \times 2}$

$\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$

---

The value of $\cos 15^\circ$ is $\frac{\sqrt{6} + \sqrt{2}}{4}$.

The final answer is $\boxed{\frac{\sqrt{6} + \sqrt{2}}{4}}$.

---

Alternate Solution:

Using $15^\circ = 60^\circ - 45^\circ$:

Let $A = 60^\circ$ and $B = 45^\circ$. Then $A - B = 60^\circ - 45^\circ = 15^\circ$.

We need the trigonometric values for $60^\circ$ and $45^\circ$:

$\cos 60^\circ = \frac{1}{2}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

---

Substitute these values into the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$:

$\cos 15^\circ = \cos(60^\circ - 45^\circ) = \cos 60^\circ \cos 45^\circ + \sin 60^\circ \sin 45^\circ$

$\cos 15^\circ = \left(\frac{1}{2}\right) \left(\frac{1}{\sqrt{2}}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{\sqrt{2}}\right)$

---

Multiply the terms:

$\cos 15^\circ = \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}$

---

Combine the fractions:

$\cos 15^\circ = \frac{1 + \sqrt{3}}{2\sqrt{2}}$

---

Rationalize the denominator:

$\cos 15^\circ = \frac{1 + \sqrt{3}}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\cos 15^\circ = \frac{(1 + \sqrt{3})\sqrt{2}}{2 \times 2}$

$\cos 15^\circ = \frac{\sqrt{2} + \sqrt{6}}{4}$

---

This gives the same result as the previous method.

To Prove:

$\sin 2x = 2 \sin x \cos x$

---

Proof:

We start with the angle addition formula for sine, which is a known trigonometric identity:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$

---

We want to find the expression for $\sin 2x$. We can write $2x$ as the sum of $x$ and $x$.

So, let $A = x$ and $B = x$ in the angle addition formula.

---

Substitute $A=x$ and $B=x$ into the formula:

$\sin(x + x) = \sin x \cos x + \cos x \sin x$

---

Simplify the left side of the equation:

$\sin(x + x) = \sin 2x$

---

Simplify the right side of the equation. The terms $\sin x \cos x$ and $\cos x \sin x$ are the same.

$\sin x \cos x + \cos x \sin x = \sin x \cos x + \sin x \cos x = 2 \sin x \cos x$

---

Equating the simplified left and right sides, we get:

$\sin 2x = 2 \sin x \cos x$

---

This proves the double angle identity for sine.

Given Equation:

$\tan x = -\frac{1}{\sqrt{3}}$

---

To Find:

The general solution of the equation.

---

Solution:

We are asked to find the general solution of the equation $\tan x = -\frac{1}{\sqrt{3}}$.

---

First, we need to find an angle $y$ such that $\tan y = -\frac{1}{\sqrt{3}}$.

We know that $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

---

Since $\tan x$ is negative, the angle $x$ must lie in the second or fourth quadrant.

The principal value for $\tan x$ in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$ would be $-\frac{\pi}{6}$.

$\tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$.

---

Alternatively, we can find an angle in the interval $[0, \pi)$. The angle in the second quadrant with reference angle $\frac{\pi}{6}$ is $\pi - \frac{\pi}{6}$.

$\pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}$.

$\tan\left(\frac{5\pi}{6}\right) = \tan\left(\pi - \frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$.

---

So, we have $\tan x = -\frac{1}{\sqrt{3}} = \tan\left(\frac{5\pi}{6}\right)$.

---

The general solution for the equation $\tan x = \tan y$ is given by $x = n\pi + y$, where $n \in \mathbb{Z}$ (the set of all integers).

---

Using $y = \frac{5\pi}{6}$, the general solution is:

---

The final answer is $\boxed{x = n\pi + \frac{5\pi}{6}, n \in \mathbb{Z}}$.

Given:

Length of the minute hand (which is the radius of the circle), $r = 1.5 \text{ cm}$.

Time elapsed = 40 minutes.

Value of $\pi = 3.14$.

---

To Find:

The distance the tip of the minute hand moves in 40 minutes.

---

Solution:

The tip of the minute hand moves along the circumference of the circle formed by its rotation. The distance it moves is the length of the arc subtended by the angle swept by the minute hand at the centre of the clock in 40 minutes.

---

The minute hand completes one full revolution ($360^\circ$) in 60 minutes.

So, the angle swept by the minute hand in 60 minutes is $360^\circ$ or $2\pi$ radians.

---

Angle swept by the minute hand in 1 minute:

Angle per minute $= \frac{360^\circ}{60 \text{ minutes}} = 6^\circ/\text{minute}$.

---

Angle swept by the minute hand in 40 minutes:

Angle in degrees $= 40 \text{ minutes} \times 6^\circ/\text{minute} = 240^\circ$.

---

To find the arc length, we need the angle in radians. We convert $240^\circ$ to radians using the conversion factor $\frac{\pi}{180^\circ}$.

Simplify the fraction:

$\theta = \frac{240\pi}{180} = \frac{24\pi}{18} = \frac{4\pi}{3}$ radians.

---

The distance moved by the tip of the minute hand (arc length $l$) is given by the formula:

$l = r\theta$

where $r$ is the radius and $\theta$ is the angle in radians.

---

Substitute the given values of $r = 1.5 \text{ cm}$ and $\theta = \frac{4\pi}{3}$ radians into the formula:

$l = 1.5 \text{ cm} \times \frac{4\pi}{3}$

Write 1.5 as a fraction: $1.5 = \frac{15}{10} = \frac{3}{2}$.

$l = \frac{3}{2} \times \frac{4\pi}{3} \text{ cm}$

---

Cancel out the common factors:

$l = \frac{\cancel{3}}{2} \times \frac{\cancel{4}^{\;2}\pi}{\cancel{3}} \text{ cm}$

$l = 2\pi \text{ cm}$.

---

Now, substitute the given value of $\pi = 3.14$:

$l = 2 \times 3.14 \text{ cm}$

$l = 6.28 \text{ cm}$.

---

The distance the tip of the minute hand moves in 40 minutes is $6.28 \text{ cm}$.

The final answer is $\boxed{6.28 \text{ cm}}$.

Given:

$\tan x = \frac{3}{4}$

The angle $x$ lies in the third quadrant.

---

To Find:

The values of $\sin x$ and $\cos x$.

---

Solution:

We are given $\tan x = \frac{3}{4}$. Since $x$ is in the third quadrant, $\tan x$ is positive, which is consistent with the given value.

---

We use the trigonometric identity relating $\tan x$ and $\sec x$:

---

Substitute the given value of $\tan x = \frac{3}{4}$ into the identity:

$\sec^2 x = 1 + \left(\frac{3}{4}\right)^2$

$\sec^2 x = 1 + \frac{9}{16}$

To add 1 and $\frac{9}{16}$, find a common denominator:

$\sec^2 x = \frac{16}{16} + \frac{9}{16}$

$\sec^2 x = \frac{16 + 9}{16}$

$\sec^2 x = \frac{25}{16}$

---

Take the square root of both sides to find $\sec x$:

$\sec x = \pm \sqrt{\frac{25}{16}}$

$\sec x = \pm \frac{5}{4}$

---

Since the angle $x$ lies in the third quadrant, the value of $\sec x$ is negative.

---

Now we can find $\cos x$ using the reciprocal identity $\cos x = \frac{1}{\sec x}$:

$\cos x = \frac{1}{-\frac{5}{4}}$

$\cos x = -\frac{4}{5}$

---

Next, we can find $\sin x$ using the identity $\tan x = \frac{\sin x}{\cos x}$, which can be rearranged as $\sin x = \tan x \times \cos x$:

Substitute the given value of $\tan x = \frac{3}{4}$ and the calculated value of $\cos x = -\frac{4}{5}$:

$\sin x = \left(\frac{3}{4}\right) \times \left(-\frac{4}{5}\right)$

Multiply the numerators and denominators:

$\sin x = -\frac{3 \times 4}{4 \times 5}$

Cancel out the common factor of 4 in the numerator and denominator:

$\sin x = -\frac{3 \times \cancel{4}}{\cancel{4} \times 5}$

$\sin x = -\frac{3}{5}$

---

Alternatively, after finding $\cos x$, we could use the identity $\sin^2 x + \cos^2 x = 1$ to find $\sin x$.

$\sin^2 x = 1 - \cos^2 x$

$\sin^2 x = 1 - \left(-\frac{4}{5}\right)^2$

$\sin^2 x = 1 - \frac{16}{25}$

$\sin^2 x = \frac{25}{25} - \frac{16}{25}$

$\sin^2 x = \frac{25 - 16}{25}$

$\sin^2 x = \frac{9}{25}$

---

Taking the square root:

$\sin x = \pm \sqrt{\frac{9}{25}}$

$\sin x = \pm \frac{3}{5}$

---

Since $x$ is in the third quadrant, $\sin x$ is negative. Therefore, $\sin x = -\frac{3}{5}$. This confirms the previous result.

---

The values of $\sin x$ and $\cos x$ are $-\frac{3}{5}$ and $-\frac{4}{5}$, respectively.

The final answer is $\boxed{\sin x = -\frac{3}{5}, \cos x = -\frac{4}{5}}$.

To Prove:

$\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$

---

Proof:

We start with the Left Hand Side (LHS) of the identity:

LHS $= \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x}$

---

We use the sum-to-product trigonometric identities:

$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

---

In the numerator, let $A = 5x$ and $B = 3x$.

Numerator: $\sin 5x + \sin 3x$

$\frac{A+B}{2} = \frac{5x+3x}{2} = \frac{8x}{2} = 4x$

$\frac{A-B}{2} = \frac{5x-3x}{2} = \frac{2x}{2} = x$

So, $\sin 5x + \sin 3x = 2 \sin(4x) \cos(x)$.

---

In the denominator, let $A = 5x$ and $B = 3x$.

Denominator: $\cos 5x + \cos 3x$

$\frac{A+B}{2} = \frac{5x+3x}{2} = 4x$

$\frac{A-B}{2} = \frac{5x-3x}{2} = x$

So, $\cos 5x + \cos 3x = 2 \cos(4x) \cos(x)$.

---

Substitute these expressions back into the LHS:

LHS $= \frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)}$

---

Assuming $\cos(x) \neq 0$ and $\cos(4x) \neq 0$, we can cancel the common terms $2$ and $\cos(x)$ from the numerator and the denominator:

LHS $= \frac{\cancel{2} \sin(4x) \cancel{\cos(x)}}{\cancel{2} \cos(4x) \cancel{\cos(x)}}$

LHS $= \frac{\sin(4x)}{\cos(4x)}$

---

Using the identity $\frac{\sin \theta}{\cos \theta} = \tan \theta$, with $\theta = 4x$, we get:

LHS $= \tan(4x)$

---

This is the Right Hand Side (RHS) of the identity.

Since LHS = RHS, the identity is proven.

$\therefore \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$

Given Equation:

$\cos x = \frac{1}{2}$

---

To Find:

The principal solutions and the general solution of the equation.

---

Finding Principal Solutions:

The principal solutions of a trigonometric equation are the solutions that lie in the interval $[0, 2\pi)$.

---

We are given $\cos x = \frac{1}{2}$.

We know that $\cos \theta = \frac{1}{2}$ for $\theta = \frac{\pi}{3}$.

So, one possible value for $x$ is $\frac{\pi}{3}$.

---

The cosine function is positive in the first quadrant and the fourth quadrant.

---

In the first quadrant, the principal solution is the reference angle itself.

---

In the fourth quadrant, the angle with the same cosine value as $\frac{\pi}{3}$ is given by $2\pi - \theta$, where $\theta = \frac{\pi}{3}$.

$x = 2\pi - \frac{\pi}{3}$

To simplify, find a common denominator:

$x = \frac{6\pi}{3} - \frac{\pi}{3} = \frac{6\pi - \pi}{3} = \frac{5\pi}{3}$

---

Both $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ lie within the principal range $[0, 2\pi)$.

---

Thus, the principal solutions of the equation $\cos x = \frac{1}{2}$ are $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$.

---

Finding General Solution:

The general solution for the equation $\cos x = \cos y$ is given by $x = 2n\pi \pm y$, where $n \in \mathbb{Z}$ (the set of all integers).

---

From our principal solutions or basic knowledge, we know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.

So, we can write the given equation as $\cos x = \cos\left(\frac{\pi}{3}\right)$.

---

Using the general solution formula with $y = \frac{\pi}{3}$, the general solution is:

---

The principal solutions are $\frac{\pi}{3}, \frac{5\pi}{3}$.

The general solution is $x = 2n\pi \pm \frac{\pi}{3}$, where $n \in \mathbb{Z}$.

The final answer is $\boxed{\text{Principal Solutions: } x = \frac{\pi}{3}, \frac{5\pi}{3}; \text{ General Solution: } x = 2n\pi \pm \frac{\pi}{3}, n \in \mathbb{Z}}$.

Given Function:

$f(x) = \sin x$

---

To Find:

The domain and range of the function $f(x) = \sin x$.

---

Solution:

The domain of a function is the set of all possible input values (the independent variable), which are the values of $x$ for which the function is defined.

---

For the function $f(x) = \sin x$, the sine of any real number can be calculated. There are no restrictions on the input values of $x$. Therefore, the domain of $\sin x$ is the set of all real numbers.

The domain can be represented as $\mathbb{R}$ or $(-\infty, \infty)$.

---

The range of a function is the set of all possible output values (the dependent variable), which are the values of $f(x)$ that result from the input values in the domain.

---

For the function $f(x) = \sin x$, the value of $\sin x$ oscillates between $-1$ and $1$, inclusive. The maximum value that $\sin x$ can take is $1$, and the minimum value is $-1$.

Therefore, the range of $\sin x$ is the closed interval $[-1, 1]$.

---

In summary:

The domain of $f(x) = \sin x$ is $\mathbb{R}$ (all real numbers).

The range of $f(x) = \sin x$ is $[-1, 1]$.

The final answer is $\boxed{\text{Domain: } \mathbb{R}, \text{ Range: } [-1, 1]}$.

To Find:

The value of $\sin 75^\circ$.

---

Solution:

We can express $75^\circ$ as the sum of two standard angles whose trigonometric values are known. For example, $75^\circ = 45^\circ + 30^\circ$.

---

We use the angle addition formula for sine:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$

---

Let $A = 45^\circ$ and $B = 30^\circ$. Then $A + B = 45^\circ + 30^\circ = 75^\circ$.

We need the trigonometric values for $45^\circ$ and $30^\circ$:

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

$\cos 45^\circ = \frac{1}{\sqrt{2}}$

$\sin 30^\circ = \frac{1}{2}$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

---

Substitute these values into the formula for $\sin(A + B)$:

$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$

---

Multiply the terms:

$\sin 75^\circ = \frac{1 \times \sqrt{3}}{\sqrt{2} \times 2} + \frac{1 \times 1}{\sqrt{2} \times 2}$

$\sin 75^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$

---

Combine the fractions since they have a common denominator:

$\sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

---

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$\sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\sin 75^\circ = \frac{(\sqrt{3} + 1)\sqrt{2}}{2 \times (\sqrt{2})^2}$

$\sin 75^\circ = \frac{\sqrt{3} \times \sqrt{2} + 1 \times \sqrt{2}}{2 \times 2}$

$\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$

---

The value of $\sin 75^\circ$ is $\frac{\sqrt{6} + \sqrt{2}}{4}$.

The final answer is $\boxed{\frac{\sqrt{6} + \sqrt{2}}{4}}$.

To Prove:

$\cos 2x = 2 \cos^2 x - 1$ and $\cos 2x = 1 - 2 \sin^2 x$.

---

Proof:

We start with the angle addition formula for cosine, which is a known trigonometric identity:

$\cos(A + B) = \cos A \cos B - \sin A \sin B$

---

To find the expression for $\cos 2x$, we can write $2x$ as the sum of $x$ and $x$.

So, let $A = x$ and $B = x$ in the angle addition formula.

---

Substitute $A=x$ and $B=x$ into the formula:

$\cos(x + x) = \cos x \cos x - \sin x \sin x$

$\cos 2x = \cos^2 x - \sin^2 x$

This is one form of the double angle identity for cosine.

---

Now, we will prove the first required identity, $\cos 2x = 2 \cos^2 x - 1$.

We use the fundamental Pythagorean identity: $\sin^2 x + \cos^2 x = 1$.

From this identity, we can express $\sin^2 x$ in terms of $\cos^2 x$:

---

Substitute the expression for $\sin^2 x$ from equation (ii) into equation (i):

$\cos 2x = \cos^2 x - (1 - \cos^2 x)$

$\cos 2x = \cos^2 x - 1 + \cos^2 x$

$\cos 2x = \cos^2 x + \cos^2 x - 1$

$\cos 2x = 2 \cos^2 x - 1$

---

Next, we will prove the second required identity, $\cos 2x = 1 - 2 \sin^2 x$.

Again, we use the fundamental Pythagorean identity: $\sin^2 x + \cos^2 x = 1$.

From this identity, we can express $\cos^2 x$ in terms of $\sin^2 x$:

---

Substitute the expression for $\cos^2 x$ from equation (iii) into equation (i):

$\cos 2x = (1 - \sin^2 x) - \sin^2 x$

$\cos 2x = 1 - \sin^2 x - \sin^2 x$

$\cos 2x = 1 - 2 \sin^2 x$

---

Combining the results, we have proven that $\cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$.

Given Equation:

$\sqrt{3} \text{cosec } x + 2 = 0$

---

To Find:

The general solution of the equation.

---

Solution:

We are asked to find the general solution of the equation $\sqrt{3} \text{cosec } x + 2 = 0$.

---

First, isolate $\text{cosec } x$:

$\sqrt{3} \text{cosec } x = -2$

$\text{cosec } x = -\frac{2}{\sqrt{3}}$

---

We know that $\text{cosec } x = \frac{1}{\sin x}$. So, we can rewrite the equation in terms of $\sin x$:

$\frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$

Taking the reciprocal of both sides:

$\sin x = -\frac{\sqrt{3}}{2}$

---

Now we need to find the general solution of the equation $\sin x = -\frac{\sqrt{3}}{2}$.

---

First, find the reference angle. We know that $\sin \theta = \frac{\sqrt{3}}{2}$ for $\theta = \frac{\pi}{3}$.

---

Since $\sin x$ is negative, the angle $x$ must lie in the third quadrant or the fourth quadrant.

---

In the third quadrant, the angle with reference angle $\frac{\pi}{3}$ is $\pi + \frac{\pi}{3}$.

$\pi + \frac{\pi}{3} = \frac{3\pi + \pi}{3} = \frac{4\pi}{3}$.

---

In the fourth quadrant, the angle with reference angle $\frac{\pi}{3}$ is $2\pi - \frac{\pi}{3}$.

$2\pi - \frac{\pi}{3} = \frac{6\pi - \pi}{3} = \frac{5\pi}{3}$.

---

The general solution for the equation $\sin x = \sin y$ is given by $x = n\pi + (-1)^n y$, where $n \in \mathbb{Z}$.

---

We can use either principal value as $y$. Let's use $y = \frac{4\pi}{3}$.

---

Alternatively, if we use $y = -\frac{\pi}{3}$ (the principal value in $(-\frac{\pi}{2}, \frac{\pi}{2})$ for $\sin x = -\frac{\sqrt{3}}{2}$), the general solution is:

This can also be written as $x = n\pi - (-1)^n \frac{\pi}{3}$, or $x = n\pi + (-1)^{n+1} \frac{\pi}{3}$. This form represents the same set of solutions as $x = n\pi + (-1)^n \frac{4\pi}{3}$.

---

The general solution of the equation $\sqrt{3} \text{cosec } x + 2 = 0$ is $x = n\pi + (-1)^n \frac{4\pi}{3}$, where $n \in \mathbb{Z}$.

The final answer is $\boxed{x = n\pi + (-1)^n \frac{4\pi}{3}, n \in \mathbb{Z}}$.

Given:

Number of revolutions made by the wheel in one minute = 360.

---

To Find:

The number of radians the wheel turns in one second.

---

Solution:

We are given that the wheel makes 360 revolutions in one minute.

---

First, convert the time from minutes to seconds. We know that 1 minute = 60 seconds.

So, the wheel makes 360 revolutions in 60 seconds.

---

Now, calculate the number of revolutions per second:

Revolutions per second $= \frac{\text{Total revolutions}}{\text{Total time in seconds}}$

Revolutions per second $= \frac{360 \text{ revolutions}}{60 \text{ seconds}}$

Revolutions per second $= \frac{360}{60} = 6$ revolutions per second.

---

Next, we need to convert the number of revolutions into radians.

One complete revolution corresponds to an angle of $360^\circ$ or $2\pi$ radians.

---

The angle turned in radians per revolution is $2\pi$ radians.

---

To find the total radians turned in one second, we multiply the number of revolutions per second by the radians per revolution:

Radians per second $= (\text{Revolutions per second}) \times (\text{Radians per revolution})$

Radians per second $= 6 \times 2\pi$ radians

Radians per second $= 12\pi$ radians.

---

The wheel turns through $12\pi$ radians in one second.

The final answer is $\boxed{12\pi \text{ radians}}$.

To Find:

The value of $\tan \frac{13\pi}{12}$.

---

Solution:

We need to find the value of $\tan \frac{13\pi}{12}$. We can rewrite the angle $\frac{13\pi}{12}$ as a sum or difference of standard angles.

$\frac{13\pi}{12} = \frac{12\pi + \pi}{12} = \frac{12\pi}{12} + \frac{\pi}{12} = \pi + \frac{\pi}{12}$.

---

Using the property $\tan(\pi + \theta) = \tan \theta$, we have:

$\tan \frac{13\pi}{12} = \tan\left(\pi + \frac{\pi}{12}\right) = \tan\left(\frac{\pi}{12}\right)$

---

Now we need to find the value of $\tan\left(\frac{\pi}{12}\right)$. We can write $\frac{\pi}{12}$ as the difference of two standard angles:

$\frac{\pi}{12}$ radians corresponds to $\frac{180^\circ}{12} = 15^\circ$.

We can express $15^\circ$ as $45^\circ - 30^\circ$ or $60^\circ - 45^\circ$. In radians, this is $\frac{\pi}{4} - \frac{\pi}{6}$ or $\frac{\pi}{3} - \frac{\pi}{4}$.

Let's use $\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$.

---

We use the angle difference formula for tangent:

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

---

Let $A = \frac{\pi}{4}$ and $B = \frac{\pi}{6}$.

We need the tangent values for these angles:

$\tan \frac{\pi}{4} = \tan 45^\circ = 1$

$\tan \frac{\pi}{6} = \tan 30^\circ = \frac{1}{\sqrt{3}}$

---

Substitute these values into the formula for $\tan(A - B)$:

$\tan \frac{\pi}{12} = \tan\left(\frac{\pi}{4} - \frac{\pi}{6}\right) = \frac{\tan \frac{\pi}{4} - \tan \frac{\pi}{6}}{1 + \tan \frac{\pi}{4} \tan \frac{\pi}{6}}$

---

Simplify the numerator and the denominator:

Numerator: $1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}$

Denominator: $1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}$

---

Substitute these back into the fraction:

$\tan \frac{\pi}{12} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}$

$\tan \frac{\pi}{12} = \frac{\sqrt{3} - 1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3} + 1}$

Cancel out the common factor $\sqrt{3}$:

$\tan \frac{\pi}{12} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

---

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - 1$:

$\tan \frac{\pi}{12} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}$

$\tan \frac{\pi}{12} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2}$

Expand the numerator using $(a-b)^2 = a^2 - 2ab + b^2$ and simplify the denominator:

$\tan \frac{\pi}{12} = \frac{(\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2}{3 - 1}$

$\tan \frac{\pi}{12} = \frac{3 - 2\sqrt{3} + 1}{2}$

$\tan \frac{\pi}{12} = \frac{4 - 2\sqrt{3}}{2}$

---

Factor out 2 from the numerator and simplify:

$\tan \frac{\pi}{12} = \frac{2(2 - \sqrt{3})}{2}$

$\tan \frac{\pi}{12} = 2 - \sqrt{3}$

---

Since $\tan \frac{13\pi}{12} = \tan \frac{\pi}{12}$, the value is $2 - \sqrt{3}$.

The final answer is $\boxed{2 - \sqrt{3}}$.

To Prove:

$\cos(x-y) = \cos x \cos y + \sin x \sin y$

---

Proof:

We start with the angle addition formula for cosine, which is a fundamental trigonometric identity:

$\cos(A + B) = \cos A \cos B - \sin A \sin B$

---

To derive the formula for $\cos(x-y)$, we can substitute $A=x$ and $B=-y$ into the angle addition formula for cosine.

---

Substitute $A=x$ and $B=-y$ into the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:

$\cos(x + (-y)) = \cos x \cos(-y) - \sin x \sin(-y)$

---

Simplify the left side of the equation:

$\cos(x + (-y)) = \cos(x - y)$

---

Now, consider the right side of the equation: $\cos x \cos(-y) - \sin x \sin(-y)$.

We use the properties of cosine and sine for negative angles:

$\cos(-\theta) = \cos \theta$

$\sin(-\theta) = -\sin \theta$

---

Applying these properties with $\theta = y$:

$\cos(-y) = \cos y$

$\sin(-y) = -\sin y$

---

Substitute these results back into the right side of the equation for $\cos(x + (-y))$:

Right Hand Side $= \cos x (\cos y) - \sin x (-\sin y)$

Right Hand Side $= \cos x \cos y - (-\sin x \sin y)$

Right Hand Side $= \cos x \cos y + \sin x \sin y$

---

Equating the simplified left and right sides, we get:

$\cos(x - y) = \cos x \cos y + \sin x \sin y$

---

This proves the angle difference identity for cosine.

Given Equation:

$\sin 2x - \sin 4x + \sin 6x = 0$

---

To Find:

The general solution of the equation.

---

Solution:

We can rearrange the terms to group $\sin 6x$ and $\sin 2x$ together:

$(\sin 6x + \sin 2x) - \sin 4x = 0$

---

We use the sum-to-product identity for sine: $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$.

Let $A = 6x$ and $B = 2x$.

$\frac{A+B}{2} = \frac{6x+2x}{2} = \frac{8x}{2} = 4x$

$\frac{A-B}{2} = \frac{6x-2x}{2} = \frac{4x}{2} = 2x$

So, $\sin 6x + \sin 2x = 2 \sin(4x) \cos(2x)$.

---

Substitute this back into the equation:

---

Now, factor out the common term $\sin 4x$:

$\sin 4x (2 \cos 2x - 1) = 0$

---

This equation is satisfied if either factor is equal to zero.

Case 1: $\sin 4x = 0$

Case 2: $2 \cos 2x - 1 = 0$

---

Case 1: $\sin 4x = 0$

The general solution for $\sin \theta = 0$ is $\theta = n\pi$, where $n \in \mathbb{Z}$.

Here, $\theta = 4x$. So,

$4x = n\pi$

Divide by 4 to solve for $x$:

This is one part of the general solution.

---

Case 2: $2 \cos 2x - 1 = 0$

Isolate $\cos 2x$:

$2 \cos 2x = 1$

$\cos 2x = \frac{1}{2}$

---

We need to find the general solution for $\cos \theta = \frac{1}{2}$.

We know that $\cos \frac{\pi}{3} = \frac{1}{2}$.

The general solution for $\cos \theta = \cos y$ is $\theta = 2m\pi \pm y$, where $m \in \mathbb{Z}$.

Here, $\theta = 2x$ and $y = \frac{\pi}{3}$. So,

$2x = 2m\pi \pm \frac{\pi}{3}$

Divide by 2 to solve for $x$:

$x = m\pi \pm \frac{\pi}{6}$

This is the other part of the general solution.

---

Combining the solutions from both cases, the general solution of the given equation is:

$x = \frac{n\pi}{4}$ or $x = m\pi \pm \frac{\pi}{6}$, where $n, m \in \mathbb{Z}$.

The final answer is $\boxed{x = \frac{n\pi}{4} \text{ or } x = m\pi \pm \frac{\pi}{6}, \text{ where } n, m \in \mathbb{Z}}$.

Given:

Two circles have arcs of the same length.

Angle subtended by the arc at the centre of the first circle, $\theta_1 = 60^\circ$.

Angle subtended by the arc at the centre of the second circle, $\theta_2 = 75^\circ$.

---

To Find:

The ratio of their radii, $\frac{r_1}{r_2}$.

---

Solution:

Let $r_1$ and $r_2$ be the radii of the first and second circles, respectively.

Let $l_1$ and $l_2$ be the lengths of the arcs in the first and second circles, respectively.

---

We are given that the arc lengths are the same, so $l_1 = l_2$. Let this common length be $l$.

$l_1 = l$

$l_2 = l$

---

The relation between arc length ($l$), radius ($r$), and angle ($\theta$) subtended at the centre (in radians) is $l = r\theta$.

Before using this formula, we need to convert the given angles from degrees to radians.

The conversion formula is Radians = Degrees $\times \frac{\pi}{180}$.

---

Convert $\theta_1$ to radians:

$\theta_1 = 60^\circ \times \frac{\pi}{180^\circ} = \frac{60\pi}{180} = \frac{\pi}{3}$ radians.

---

Convert $\theta_2$ to radians:

$\theta_2 = 75^\circ \times \frac{\pi}{180^\circ} = \frac{75\pi}{180}$.

Simplify the fraction $\frac{75}{180}$ by dividing both by 15 (since $75 = 5 \times 15$ and $180 = 12 \times 15$).

$\theta_2 = \frac{5\pi}{12}$ radians.

---

Now, apply the formula $l = r\theta$ to both circles:

For the first circle: $l_1 = r_1 \theta_1 \implies l = r_1 \left(\frac{\pi}{3}\right)$

For the second circle: $l_2 = r_2 \theta_2 \implies l = r_2 \left(\frac{5\pi}{12}\right)$

---

We need to find the ratio $\frac{r_1}{r_2}$. We can express $r_1$ and $r_2$ in terms of $l$ from equations (i) and (ii):

From (i), $r_1 = \frac{l}{\frac{\pi}{3}} = \frac{3l}{\pi}$.

From (ii), $r_2 = \frac{l}{\frac{5\pi}{12}} = \frac{12l}{5\pi}$.

---

Now, find the ratio $\frac{r_1}{r_2}$:

$\frac{r_1}{r_2} = \frac{\frac{3l}{\pi}}{\frac{12l}{5\pi}}$

$\frac{r_1}{r_2} = \frac{3l}{\pi} \times \frac{5\pi}{12l}$

---

Cancel out common terms $l$ and $\pi$:

$\frac{r_1}{r_2} = \frac{3\cancel{l}}{\cancel{\pi}} \times \frac{5\cancel{\pi}}{12\cancel{l}}$

$\frac{r_1}{r_2} = \frac{3 \times 5}{12}$

Simplify the fraction by canceling out a factor of 3:

$\frac{r_1}{r_2} = \frac{\cancel{3}^{\;1} \times 5}{\cancel{12}_{\;4}}$

$\frac{r_1}{r_2} = \frac{1 \times 5}{4} = \frac{5}{4}$

---

The ratio of the radii is $5:4$.

The final answer is $\boxed{5:4}$.

To Prove:

$\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \text{cot } 3x$

---

Proof:

We start with the Left Hand Side (LHS) of the identity:

LHS $= \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}$

---

We can rearrange the terms in the numerator and denominator to group the terms which, when their arguments are averaged, give the argument of the middle term ($3x$).

LHS $= \frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x}$

---

We use the sum-to-product trigonometric identities:

$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

---

Apply the identities to the terms in the numerator $(\cos 4x + \cos 2x)$ with $A = 4x$ and $B = 2x$:

$\frac{A+B}{2} = \frac{4x+2x}{2} = \frac{6x}{2} = 3x$

$\frac{A-B}{2} = \frac{4x-2x}{2} = \frac{2x}{2} = x$

So, $\cos 4x + \cos 2x = 2 \cos(3x) \cos(x)$.

---

The numerator becomes: $2 \cos(3x) \cos(x) + \cos 3x$.

Factor out the common term $\cos 3x$ from the numerator:

Numerator $= \cos 3x (2 \cos x + 1)$.

---

Apply the identities to the terms in the denominator $(\sin 4x + \sin 2x)$ with $A = 4x$ and $B = 2x$:

$\frac{A+B}{2} = \frac{4x+2x}{2} = 3x$

$\frac{A-B}{2} = \frac{4x-2x}{2} = x$

So, $\sin 4x + \sin 2x = 2 \sin(3x) \cos(x)$.

---

The denominator becomes: $2 \sin(3x) \cos(x) + \sin 3x$.

Factor out the common term $\sin 3x$ from the denominator:

Denominator $= \sin 3x (2 \cos x + 1)$.

---

Substitute the factored expressions back into the LHS:

LHS $= \frac{\cos 3x (2 \cos x + 1)}{\sin 3x (2 \cos x + 1)}$

---

Assuming $2 \cos x + 1 \neq 0$, we can cancel the common factor $(2 \cos x + 1)$ from the numerator and the denominator:

LHS $= \frac{\cos 3x \cancel{(2 \cos x + 1)}}{\sin 3x \cancel{(2 \cos x + 1)}}$

LHS $= \frac{\cos 3x}{\sin 3x}$

---

Using the quotient identity $\frac{\cos \theta}{\sin \theta} = \cot \theta$, with $\theta = 3x$, we get:

LHS $= \cot 3x$

---

This is the Right Hand Side (RHS) of the identity.

Since LHS = RHS, the identity is proven.

$\therefore \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \text{cot } 3x$

To Prove:

$\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$

---

Proof:

We start with the Left Hand Side (LHS) of the identity:

LHS $= \sin x + \sin 3x + \sin 5x + \sin 7x$

---

We group the terms strategically to apply the sum-to-product identities. Let's group $(\sin 7x + \sin x)$ and $(\sin 5x + \sin 3x)$ because the average of their arguments leads to $4x$, which appears in the RHS.

LHS $= (\sin 7x + \sin x) + (\sin 5x + \sin 3x)$

---

We use the sum-to-product identity: $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$.

---

Apply the identity to the first group $(\sin 7x + \sin x)$ with $A = 7x$ and $B = x$:

$\frac{A+B}{2} = \frac{7x+x}{2} = \frac{8x}{2} = 4x$

$\frac{A-B}{2} = \frac{7x-x}{2} = \frac{6x}{2} = 3x$

So, $\sin 7x + \sin x = 2 \sin(4x) \cos(3x)$.

---

Apply the identity to the second group $(\sin 5x + \sin 3x)$ with $A = 5x$ and $B = 3x$:

$\frac{A+B}{2} = \frac{5x+3x}{2} = \frac{8x}{2} = 4x$

$\frac{A-B}{2} = \frac{5x-3x}{2} = \frac{2x}{2} = x$

So, $\sin 5x + \sin 3x = 2 \sin(4x) \cos(x)$.

---

Substitute these results back into the LHS expression:

LHS $= (2 \sin 4x \cos 3x) + (2 \sin 4x \cos x)$

---

Factor out the common term $2 \sin 4x$ from both terms:

LHS $= 2 \sin 4x (\cos 3x + \cos x)$

---

Now, we use the sum-to-product identity for cosine: $\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$.

Apply the identity to the term $(\cos 3x + \cos x)$ with $A = 3x$ and $B = x$:

$\frac{A+B}{2} = \frac{3x+x}{2} = \frac{4x}{2} = 2x$

$\frac{A-B}{2} = \frac{3x-x}{2} = \frac{2x}{2} = x$

So, $\cos 3x + \cos x = 2 \cos(2x) \cos(x)$.

---

Substitute this result back into the LHS expression:

LHS $= 2 \sin 4x (2 \cos 2x \cos x)$

---

Multiply the terms:

LHS $= 2 \times 2 \times \cos x \times \cos 2x \times \sin 4x$

LHS $= 4 \cos x \cos 2x \sin 4x$

---

This is the Right Hand Side (RHS) of the identity.

Since LHS = RHS, the identity is proven.

$\therefore \sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$

Given Equation:

$\cos 3x + \cos x - \cos 2x = 0$

---

To Find:

The general solution of the equation.

---

Solution:

We are asked to find the general solution of the equation $\cos 3x + \cos x - \cos 2x = 0$.

---

We can group the terms $\cos 3x$ and $\cos x$ together:

$(\cos 3x + \cos x) - \cos 2x = 0$

---

We use the sum-to-product trigonometric identity:

$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

---

Apply the identity to the term $(\cos 3x + \cos x)$ with $A = 3x$ and $B = x$:

$\frac{A+B}{2} = \frac{3x+x}{2} = \frac{4x}{2} = 2x$

$\frac{A-B}{2} = \frac{3x-x}{2} = \frac{2x}{2} = x$

So, $\cos 3x + \cos x = 2 \cos(2x) \cos(x)$.

---

Substitute this result back into the given equation:

---

Now, factor out the common term $\cos 2x$:

$\cos 2x (2 \cos x - 1) = 0$

---

This equation is satisfied if either factor is equal to zero.

Case 1: $\cos 2x = 0$

Case 2: $2 \cos x - 1 = 0$

---

Case 1: $\cos 2x = 0$

The general solution for $\cos \theta = 0$ is $\theta = (2n+1)\frac{\pi}{2}$, where $n \in \mathbb{Z}$.

Here, $\theta = 2x$. So,

$2x = (2n+1)\frac{\pi}{2}$

Divide both sides by 2:

This is one set of solutions.

---

Case 2: $2 \cos x - 1 = 0$

Isolate $\cos x$:

$2 \cos x = 1$

$\cos x = \frac{1}{2}$

---

We know that $\cos \frac{\pi}{3} = \frac{1}{2}$.

So, the equation is $\cos x = \cos \frac{\pi}{3}$.

The general solution for $\cos x = \cos y$ is $x = 2m\pi \pm y$, where $m \in \mathbb{Z}$.

Here, $y = \frac{\pi}{3}$. So,

This is the other set of solutions.

---

The general solution of the equation $\cos 3x + \cos x - \cos 2x = 0$ is the union of the solutions from Case 1 and Case 2.

The general solution is $x = \frac{(2n+1)\pi}{4}$ or $x = 2m\pi \pm \frac{\pi}{3}$, where $n, m \in \mathbb{Z}$.

The final answer is $\boxed{x = \frac{(2n+1)\pi}{4} \text{ or } x = 2m\pi \pm \frac{\pi}{3}, \text{ where } n, m \in \mathbb{Z}}$.

To Prove:

$\tan(x+y+z) = \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan x \tan y - \tan y \tan z - \tan z \tan x}$

---

Proof:

We start with the Left Hand Side (LHS) of the identity:

LHS $= \tan(x+y+z)$

---

We can group the terms in the argument of the tangent function. Let $A = x+y$ and $B = z$. Then we can use the angle addition formula for tangent:

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

---

Substitute $A = x+y$ and $B = z$ into the formula:

$\tan(x+y+z) = \tan((x+y) + z) = \frac{\tan(x+y) + \tan z}{1 - \tan(x+y) \tan z}$

---

Now, we need to find the expression for $\tan(x+y)$ using the angle addition formula for tangent with $A = x$ and $B = y$:

---

Substitute the expression for $\tan(x+y)$ from equation (ii) into equation (i):

LHS $= \frac{\left(\frac{\tan x + \tan y}{1 - \tan x \tan y}\right) + \tan z}{1 - \left(\frac{\tan x + \tan y}{1 - \tan x \tan y}\right) \tan z}$

---

Now, we simplify the numerator and the denominator by finding a common denominator in each.

Numerator: $\frac{\tan x + \tan y}{1 - \tan x \tan y} + \tan z = \frac{\tan x + \tan y + \tan z (1 - \tan x \tan y)}{1 - \tan x \tan y}$

Numerator $= \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan x \tan y}$

---

Denominator: $1 - \left(\frac{\tan x + \tan y}{1 - \tan x \tan y}\right) \tan z = 1 - \frac{(\tan x + \tan y) \tan z}{1 - \tan x \tan y}$

Denominator $= \frac{(1 - \tan x \tan y) - (\tan x + \tan y) \tan z}{1 - \tan x \tan y}$

Denominator $= \frac{1 - \tan x \tan y - \tan x \tan z - \tan y \tan z}{1 - \tan x \tan y}$

---

Substitute the simplified numerator and denominator back into the LHS expression:

LHS $= \frac{\frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan x \tan y}}{\frac{1 - \tan x \tan y - \tan x \tan z - \tan y \tan z}{1 - \tan x \tan y}}$

---

Multiply the numerator by the reciprocal of the denominator:

LHS $= \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan x \tan y} \times \frac{1 - \tan x \tan y}{1 - \tan x \tan y - \tan x \tan z - \tan y \tan z}$

---

Cancel out the common term $(1 - \tan x \tan y)$ from the numerator and the denominator, assuming $1 - \tan x \tan y \neq 0$:

LHS $= \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan x \tan y - \tan x \tan z - \tan y \tan z}$

---

Rearrange the terms in the denominator to match the required form:

LHS $= \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan x \tan y - \tan y \tan z - \tan z \tan x}$

---

This is the Right Hand Side (RHS) of the identity.

Since LHS = RHS, the identity is proven.

$\therefore \tan(x+y+z) = \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan x \tan y - \tan y \tan z - \tan z \tan x}$

Given Equation:

$2 \cos^2 x + 3 \sin x = 0$

---

To Find:

The general solution of the equation.

---

Solution:

We are asked to find the general solution of the equation $2 \cos^2 x + 3 \sin x = 0$.

---

We can rewrite the equation in terms of a single trigonometric function. We use the fundamental trigonometric identity $\cos^2 x + \sin^2 x = 1$, which implies $\cos^2 x = 1 - \sin^2 x$.

---

Substitute $\cos^2 x = 1 - \sin^2 x$ into the given equation:

Distribute the 2:

$2 - 2 \sin^2 x + 3 \sin x = 0$

---

Rearrange the terms to form a quadratic equation in terms of $\sin x$:

$-2 \sin^2 x + 3 \sin x + 2 = 0$

Multiply the equation by $-1$ to make the leading coefficient positive:

$2 \sin^2 x - 3 \sin x - 2 = 0$

---

Let $y = \sin x$. The equation becomes a quadratic equation in $y$:

$2y^2 - 3y - 2 = 0$

---

We can solve this quadratic equation by factoring. We look for two numbers that multiply to $(2)(-2) = -4$ and add up to $-3$. These numbers are $-4$ and $1$.

Rewrite the middle term using these numbers:

$2y^2 - 4y + y - 2 = 0$

Group terms and factor:

$(2y^2 - 4y) + (y - 2) = 0$

$2y(y - 2) + 1(y - 2) = 0$

Factor out the common term $(y - 2)$:

$(2y + 1)(y - 2) = 0$

---

This equation implies that either $2y + 1 = 0$ or $y - 2 = 0$. Substitute $\sin x$ back for $y$ to get two separate trigonometric equations:

Case 1: $2 \sin x + 1 = 0 \implies \sin x = -\frac{1}{2}$

Case 2: $\sin x - 2 = 0 \implies \sin x = 2$

---

Case 1: $\sin x = -\frac{1}{2}$

We need to find the general solution for $\sin x = -\frac{1}{2}$.

We know that $\sin \theta = \frac{1}{2}$ for $\theta = \frac{\pi}{6}$. Since $\sin x$ is negative, $x$ lies in the third or fourth quadrant.

The principal value of $x$ such that $\sin x = -\frac{1}{2}$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is $-\frac{\pi}{6}$.

So, we have $\sin x = \sin\left(-\frac{\pi}{6}\right)$.

---

The general solution for the equation $\sin x = \sin y$ is given by $x = n\pi + (-1)^n y$, where $n \in \mathbb{Z}$ (the set of all integers).

Using $y = -\frac{\pi}{6}$, the general solution for Case 1 is:

This can be written as $x = n\pi - (-1)^n \frac{\pi}{6}$, or $x = n\pi + (-1)^{n+1} \frac{\pi}{6}$, where $n \in \mathbb{Z}$.

---

Case 2: $\sin x = 2$

The range of the sine function is $[-1, 1]$. This means that the value of $\sin x$ must be between $-1$ and $1$, inclusive.

Since $2$ is outside this range (specifically, $2 > 1$), there is no real value of $x$ for which $\sin x = 2$.

Thus, this case yields no solutions.

---

The general solution of the original equation $2 \cos^2 x + 3 \sin x = 0$ is given only by the solutions from Case 1.

The general solution is $x = n\pi + (-1)^{n+1} \frac{\pi}{6}$, where $n \in \mathbb{Z}$.

The final answer is $\boxed{x = n\pi + (-1)^{n+1} \frac{\pi}{6}, n \in \mathbb{Z}}$.

Given:

Radius of the circular wire, $r_{wire} = 3 \text{ cm}$.

Radius of the hoop, $r_{hoop} = 48 \text{ cm}$.

---

To Find:

The angle in degrees subtended by the wire at the centre of the hoop.

---

Solution:

When the circular wire is cut and bent to lie along the circumference of the hoop, the length of the wire becomes the length of the arc on the hoop.

---

The length of the circular wire is equal to its circumference. The circumference of a circle with radius $r$ is $2\pi r$.

Length of the wire $= \text{Circumference of the circular wire} = 2\pi r_{wire}$.

Length of the wire $= 2\pi (3 \text{ cm}) = 6\pi \text{ cm}$.

---

This length of the wire is the length of the arc ($l$) on the hoop.

Arc length on the hoop, $l = 6\pi \text{ cm}$.

---

The radius of the hoop is $r_{hoop} = 48 \text{ cm}$.

---

We use the relation between the arc length ($l$), radius ($r$), and the angle ($\theta$) subtended at the centre (in radians): $l = r\theta$.

Here, $r = r_{hoop} = 48 \text{ cm}$ and $l = 6\pi \text{ cm}$. We need to find the angle $\theta$ in radians first.

Rearranging the formula, we get $\theta = \frac{l}{r}$.

---

Substitute the values of $l$ and $r_{hoop}$:

Simplify the fraction:

$\theta = \frac{6\pi}{48} = \frac{\pi}{8}$ radians.

---

The question asks for the angle in degrees. We convert the angle from radians to degrees using the conversion factor $\frac{180^\circ}{\pi}$.

Angle in degrees $= \theta \times \frac{180^\circ}{\pi}$

---

Substitute the value of $\theta = \frac{\pi}{8}$ radians:

Angle in degrees $= \frac{\pi}{8} \times \frac{180^\circ}{\pi}$

---

Cancel out the common factor $\pi$:

Angle in degrees $= \frac{\cancel{\pi}}{8} \times \frac{180^\circ}{\cancel{\pi}}$

Angle in degrees $= \frac{180}{8}^\circ$

---

Simplify the fraction $\frac{180}{8}$. We can divide both by 4:

$\frac{180}{8} = \frac{45}{2} = 22.5$.

Angle in degrees $= 22.5^\circ$.

---

The angle subtended at the centre of the hoop is $22.5^\circ$.

The final answer is $\boxed{22.5^\circ}$.

To Prove:

$\sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \frac{3\pi}{5} \sin \frac{4\pi}{5} = \frac{5}{16}$

---

Proof:

We start with the Left Hand Side (LHS) of the identity:

LHS $= \sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \frac{3\pi}{5} \sin \frac{4\pi}{5}$

---

We use the trigonometric identity $\sin(\pi - \theta) = \sin \theta$.

For the term $\sin \frac{4\pi}{5}$, we have $\frac{4\pi}{5} = \pi - \frac{\pi}{5}$.

$\sin \frac{4\pi}{5} = \sin\left(\pi - \frac{\pi}{5}\right) = \sin \frac{\pi}{5}$

---

For the term $\sin \frac{3\pi}{5}$, we have $\frac{3\pi}{5} = \pi - \frac{2\pi}{5}$.

$\sin \frac{3\pi}{5} = \sin\left(\pi - \frac{2\pi}{5}\right) = \sin \frac{2\pi}{5}$

---

Substitute these results back into the LHS expression:

LHS $= \left(\sin \frac{\pi}{5}\right) \left(\sin \frac{2\pi}{5}\right) \left(\sin \frac{2\pi}{5}\right) \left(\sin \frac{\pi}{5}\right)$

LHS $= \left(\sin \frac{\pi}{5}\right)^2 \left(\sin \frac{2\pi}{5}\right)^2$

LHS $= \left(\sin \frac{\pi}{5} \sin \frac{2\pi}{5}\right)^2$

---

Now we need to evaluate the product $\sin \frac{\pi}{5} \sin \frac{2\pi}{5}$.

The angles are $\frac{\pi}{5}$ radians ($36^\circ$) and $\frac{2\pi}{5}$ radians ($72^\circ$).

We use the known values of $\sin 36^\circ$ and $\sin 72^\circ$:

$\sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$

$\sin 72^\circ = \sin(90^\circ - 18^\circ) = \cos 18^\circ = \frac{\sqrt{10 + 2\sqrt{5}}}{4}$

---

Calculate the product $\sin \frac{\pi}{5} \sin \frac{2\pi}{5} = \sin 36^\circ \sin 72^\circ$:

$\sin 36^\circ \sin 72^\circ = \left(\frac{\sqrt{10 - 2\sqrt{5}}}{4}\right) \left(\frac{\sqrt{10 + 2\sqrt{5}}}{4}\right)$

$= \frac{\sqrt{(10 - 2\sqrt{5})(10 + 2\sqrt{5})}}{16}$

Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:

$= \frac{\sqrt{10^2 - (2\sqrt{5})^2}}{16}$

$= \frac{\sqrt{100 - (4 \times 5)}}{16}$

$= \frac{\sqrt{100 - 20}}{16}$

$= \frac{\sqrt{80}}{16}$

Simplify $\sqrt{80}$: $\sqrt{80} = \sqrt{16 \times 5} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5}$.

$= \frac{4\sqrt{5}}{16}$

Cancel out the common factor 4:

$= \frac{\cancel{4}\sqrt{5}}{\cancel{16}_{\;4}}$

$= \frac{\sqrt{5}}{4}$

---

Now substitute this product back into the expression for the LHS:

LHS $= \left(\sin \frac{\pi}{5} \sin \frac{2\pi}{5}\right)^2 = \left(\frac{\sqrt{5}}{4}\right)^2$

LHS $= \frac{(\sqrt{5})^2}{4^2} = \frac{5}{16}$

---

This is the Right Hand Side (RHS) of the identity.

Since LHS = RHS, the identity is proven.

$\therefore \sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \frac{3\pi}{5} \sin \frac{4\pi}{5} = \frac{5}{16}$

Given:

$\tan x = \frac{3}{4}$

The angle $x$ lies in the third quadrant, i.e., $\pi < x < \frac{3\pi}{2}$.

---

To Find:

The values of $\sin \frac{x}{2}, \cos \frac{x}{2}, \tan \frac{x}{2}$.

---

Solution:

First, we determine the quadrant in which $\frac{x}{2}$ lies. We are given $\pi < x < \frac{3\pi}{2}$.

Divide the inequality by 2:

$\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$

This means $\frac{x}{2}$ lies in the second quadrant.

---

In the second quadrant:

$\sin \frac{x}{2}$ is positive.

$\cos \frac{x}{2}$ is negative.

$\tan \frac{x}{2}$ is negative.

---

We need to find the value of $\cos x$. We know that $\sec^2 x = 1 + \tan^2 x$.

$\sec^2 x = 1 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{16+9}{16} = \frac{25}{16}$.

$\sec x = \pm \sqrt{\frac{25}{16}} = \pm \frac{5}{4}$.

---

Since $x$ is in the third quadrant, $\cos x$ and $\sec x$ are negative.

Therefore, $\cos x = \frac{1}{\sec x} = \frac{1}{-\frac{5}{4}} = -\frac{4}{5}$.

---

We use the half-angle identities:

$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}$

$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$

---

Calculate $\sin^2 \frac{x}{2}$:

$\sin^2 \frac{x}{2} = \frac{1 - (-\frac{4}{5})}{2} = \frac{1 + \frac{4}{5}}{2}$

Numerator: $1 + \frac{4}{5} = \frac{5}{5} + \frac{4}{5} = \frac{9}{5}$.

$\sin^2 \frac{x}{2} = \frac{\frac{9}{5}}{2} = \frac{9}{5} \times \frac{1}{2} = \frac{9}{10}$.

---

Since $\frac{x}{2}$ is in the second quadrant, $\sin \frac{x}{2}$ is positive.

$\sin \frac{x}{2} = \sqrt{\frac{9}{10}} = \frac{\sqrt{9}}{\sqrt{10}} = \frac{3}{\sqrt{10}}$.

Rationalize the denominator: $\sin \frac{x}{2} = \frac{3}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$.

---

Calculate $\cos^2 \frac{x}{2}$:

$\cos^2 \frac{x}{2} = \frac{1 + (-\frac{4}{5})}{2} = \frac{1 - \frac{4}{5}}{2}$

Numerator: $1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5}$.

$\cos^2 \frac{x}{2} = \frac{\frac{1}{5}}{2} = \frac{1}{5} \times \frac{1}{2} = \frac{1}{10}$.

---

Since $\frac{x}{2}$ is in the second quadrant, $\cos \frac{x}{2}$ is negative.

$\cos \frac{x}{2} = -\sqrt{\frac{1}{10}} = -\frac{\sqrt{1}}{\sqrt{10}} = -\frac{1}{\sqrt{10}}$.

Rationalize the denominator: $\cos \frac{x}{2} = -\frac{1}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = -\frac{\sqrt{10}}{10}$.

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Calculate $\tan \frac{x}{2}$ using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\frac{3}{\sqrt{10}}}{-\frac{1}{\sqrt{10}}}$

$\tan \frac{x}{2} = \frac{3}{\sqrt{10}} \times \left(-\frac{\sqrt{10}}{1}\right)$

$\tan \frac{x}{2} = -\frac{3 \times \cancel{\sqrt{10}}}{\cancel{\sqrt{10}} \times 1} = -3$.

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Alternatively, we can use the half-angle identity for tangent: $\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$ or $\tan \frac{x}{2} = \frac{\sin x}{1 + \cos x}$.

We know $\cos x = -\frac{4}{5}$. We need to find $\sin x$. Since $x$ is in the third quadrant and $\tan x = \frac{3}{4}$, we can form a right triangle with opposite side 3 and adjacent side 4. The hypotenuse is $\sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$.

In the third quadrant, $\sin x$ is negative. So, $\sin x = -\frac{3}{5}$.

Using $\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}$:

$\tan \frac{x}{2} = \frac{1 - (-\frac{4}{5})}{-\frac{3}{5}} = \frac{1 + \frac{4}{5}}{-\frac{3}{5}}$

$\tan \frac{x}{2} = \frac{\frac{9}{5}}{-\frac{3}{5}} = \frac{9}{5} \times \left(-\frac{5}{3}\right)$

$\tan \frac{x}{2} = -\frac{9 \times \cancel{5}}{\cancel{5} \times 3} = -\frac{9}{3} = -3$.

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Using $\tan \frac{x}{2} = \frac{\sin x}{1 + \cos x}$:

$\tan \frac{x}{2} = \frac{-\frac{3}{5}}{1 + (-\frac{4}{5})} = \frac{-\frac{3}{5}}{1 - \frac{4}{5}}$

$\tan \frac{x}{2} = \frac{-\frac{3}{5}}{\frac{1}{5}} = -\frac{3}{5} \times \frac{5}{1}$

$\tan \frac{x}{2} = -\frac{3 \times \cancel{5}}{\cancel{5} \times 1} = -3$.

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All methods give the same results.

The values are $\sin \frac{x}{2} = \frac{3\sqrt{10}}{10}$, $\cos \frac{x}{2} = -\frac{\sqrt{10}}{10}$, and $\tan \frac{x}{2} = -3$.

The final answer is $\boxed{\sin \frac{x}{2} = \frac{3\sqrt{10}}{10}, \cos \frac{x}{2} = -\frac{\sqrt{10}}{10}, \tan \frac{x}{2} = -3}$.

Given Equation:

$\sin 2x - \sin 4x + \sin 6x = 0$

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To Solve:

Find the general solution of the equation.

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Solution:

We are asked to solve the equation $\sin 2x - \sin 4x + \sin 6x = 0$, which means finding its general solution.

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We can rearrange the terms to group $\sin 6x$ and $\sin 2x$ together:

$(\sin 6x + \sin 2x) - \sin 4x = 0$

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We use the sum-to-product identity for sine: $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$.

Let $A = 6x$ and $B = 2x$.

$\frac{A+B}{2} = \frac{6x+2x}{2} = \frac{8x}{2} = 4x$

$\frac{A-B}{2} = \frac{6x-2x}{2} = \frac{4x}{2} = 2x$

So, $\sin 6x + \sin 2x = 2 \sin(4x) \cos(2x)$.

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Substitute this back into the equation:

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Now, factor out the common term $\sin 4x$:

$\sin 4x (2 \cos 2x - 1) = 0$

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This equation is satisfied if either factor is equal to zero.

Case 1: $\sin 4x = 0$

Case 2: $2 \cos 2x - 1 = 0$

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Case 1: $\sin 4x = 0$

The general solution for $\sin \theta = 0$ is $\theta = n\pi$, where $n \in \mathbb{Z}$.

Here, $\theta = 4x$. So,

$4x = n\pi$

Divide by 4 to solve for $x$:

This is one part of the general solution.

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Case 2: $2 \cos 2x - 1 = 0$

Isolate $\cos 2x$:

$2 \cos 2x = 1$

$\cos 2x = \frac{1}{2}$

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We need to find the general solution for $\cos \theta = \frac{1}{2}$.

We know that $\cos \frac{\pi}{3} = \frac{1}{2}$.

So, the equation is $\cos 2x = \cos \frac{\pi}{3}$.

The general solution for $\cos \theta = \cos y$ is $\theta = 2m\pi \pm y$, where $m \in \mathbb{Z}$.

Here, $\theta = 2x$ and $y = \frac{\pi}{3}$. So,

$2x = 2m\pi \pm \frac{\pi}{3}$

Divide by 2 to solve for $x$:

$x = m\pi \pm \frac{\pi}{6}$

This is the other part of the general solution.

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Combining the solutions from both cases, the general solution of the given equation is:

$x = \frac{n\pi}{4}$ or $x = m\pi \pm \frac{\pi}{6}$, where $n, m \in \mathbb{Z}$.

The final answer is $\boxed{x = \frac{n\pi}{4} \text{ or } x = m\pi \pm \frac{\pi}{6}, \text{ where } n, m \in \mathbb{Z}}$.

To Prove:

$\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x$

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Proof:

We start by considering the relationship between the arguments $3x$, $2x$, and $x$. We can see that $3x = 2x + x$.

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Take the tangent of both sides of the equation $3x = 2x + x$:

$\tan(3x) = \tan(2x + x)$

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Use the angle addition formula for tangent on the right side: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.

Let $A = 2x$ and $B = x$.

$\tan(2x + x) = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}$

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So, we have the equation:

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Multiply both sides of the equation by the denominator $(1 - \tan 2x \tan x)$:

$\tan 3x (1 - \tan 2x \tan x) = \tan 2x + \tan x$

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Distribute $\tan 3x$ on the left side:

$\tan 3x \times 1 - \tan 3x \times (\tan 2x \tan x) = \tan 2x + \tan x$

$\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x$

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We want to isolate the product term on one side and the difference of tangent terms on the other. Move the product term to the right side and the terms $\tan 2x$ and $\tan x$ to the left side.

Add $\tan 3x \tan 2x \tan x$ to both sides:

$\tan 3x = \tan 2x + \tan x + \tan 3x \tan 2x \tan x$

Subtract $\tan 2x$ and $\tan x$ from both sides:

$\tan 3x - \tan 2x - \tan x = \tan 3x \tan 2x \tan x$

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Rearranging the equation, we get:

$\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x$

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This proves the identity.

Given:

Central angle, $\theta = 60^\circ$.

Length of the arc, $l = 37.4 \text{ cm}$.

Value of $\pi = \frac{22}{7}$.

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To Find:

The radius of the circle, $r$.

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Solution:

The relation between the arc length ($l$), radius ($r$), and the angle ($\theta$) subtended at the centre (in radians) is given by the formula:

$l = r\theta$

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Before using this formula, we need to convert the given angle from degrees to radians.

The conversion formula is Radians = Degrees $\times \frac{\pi}{180}$.

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Convert $\theta = 60^\circ$ to radians:

Simplify the fraction:

$\theta = \frac{60\pi}{180} = \frac{\pi}{3}$ radians.

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Now, we use the formula $l = r\theta$ and substitute the given values of $l$ and $\theta$ (in radians):

$37.4 = r \times \frac{\pi}{3}$

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We need to solve for $r$. Multiply both sides by 3 and divide by $\pi$:

$r = \frac{37.4 \times 3}{\pi}$

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Substitute the given value of $\pi = \frac{22}{7}$:

$r = \frac{37.4 \times 3}{\frac{22}{7}}$

$r = 37.4 \times 3 \times \frac{7}{22}$

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We can rewrite $37.4$ as $\frac{374}{10}$.

$r = \frac{374}{10} \times 3 \times \frac{7}{22}$

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Now, simplify the expression. We can divide 374 by 22. $374 = 22 \times 17$.

$r = \frac{\cancel{374}^{\;17}}{10} \times 3 \times \frac{7}{\cancel{22}_{\;1}}$

$r = \frac{17}{10} \times 3 \times 7$

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Multiply the numbers in the numerator:

$r = \frac{17 \times 3 \times 7}{10}$

$r = \frac{17 \times 21}{10}$

$r = \frac{357}{10}$

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Perform the division:

$r = 35.7$

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The radius of the circle is $35.7 \text{ cm}$.

The final answer is $\boxed{35.7 \text{ cm}}$.

To Prove:

$(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4 \cos^2 \frac{x+y}{2}$

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Proof:

We start with the Left Hand Side (LHS) of the identity:

LHS $= (\cos x + \cos y)^2 + (\sin x - \sin y)^2$

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Expand the squared terms using the formulas $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$:

$(\cos x + \cos y)^2 = \cos^2 x + 2 \cos x \cos y + \cos^2 y$

$(\sin x - \sin y)^2 = \sin^2 x - 2 \sin x \sin y + \sin^2 y$

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Add the expanded terms:

LHS $= (\cos^2 x + 2 \cos x \cos y + \cos^2 y) + (\sin^2 x - 2 \sin x \sin y + \sin^2 y)$

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Rearrange the terms to group $\sin^2$ and $\cos^2$ terms:

LHS $= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2 \cos x \cos y - 2 \sin x \sin y$

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Use the fundamental Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ for $x$ and $y$:

$\cos^2 x + \sin^2 x = 1$

$\cos^2 y + \sin^2 y = 1$

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Substitute these values back into the LHS expression:

LHS $= 1 + 1 + 2 \cos x \cos y - 2 \sin x \sin y$

LHS $= 2 + 2 (\cos x \cos y - \sin x \sin y)$

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We use the angle addition formula for cosine: $\cos(A + B) = \cos A \cos B - \sin A \sin B$.

Let $A = x$ and $B = y$. Then $\cos x \cos y - \sin x \sin y = \cos(x + y)$.

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Substitute this into the LHS expression:

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Now we relate $\cos(x+y)$ to $\cos^2 \frac{x+y}{2}$ using the double angle identity $\cos 2\theta = 2 \cos^2 \theta - 1$.

Let $\theta = \frac{x+y}{2}$. Then $2\theta = 2 \left(\frac{x+y}{2}\right) = x+y$.

So, $\cos(x+y) = 2 \cos^2 \left(\frac{x+y}{2}\right) - 1$.

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Substitute this expression for $\cos(x+y)$ back into equation (i):

LHS $= 2 + 2 \left(2 \cos^2 \left(\frac{x+y}{2}\right) - 1\right)$

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Distribute the 2:

LHS $= 2 + 4 \cos^2 \left(\frac{x+y}{2}\right) - 2$

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Combine the constant terms:

LHS $= 4 \cos^2 \left(\frac{x+y}{2}\right)$

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This is the Right Hand Side (RHS) of the identity.

Since LHS = RHS, the identity is proven.

$\therefore (\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4 \cos^2 \frac{x+y}{2}$