Chapter 8 Introduction to Trigonometry (Additional Questions)

In a right-angled triangle, the trigonometric ratios are defined based on the ratios of the sides relative to an acute angle. For an acute angle $\theta$, the side opposite to the angle is called the perpendicular, the side adjacent to the angle is called the base, and the longest side (opposite the right angle) is the hypotenuse.

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The ratio of the side opposite to the acute angle $\theta$ to the hypotenuse is defined as the sine of $\theta$. Mathematically, this is represented as:

$\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}}$

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Therefore, the ratio of the side opposite to $\theta$ to the hypotenuse is called the Sine of $\theta$.

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The correct option is (B) Sine of $\theta$.

Given:

Triangle PQR is a right-angled triangle, right-angled at Q.

PQ = 4 cm

QR = 3 cm

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To Find:

The value of $\sin R$.

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Solution:

In the right-angled triangle PQR, the side opposite to angle R is PQ, and the side adjacent to angle R is QR.

The hypotenuse is the side opposite to the right angle Q, which is PR.

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We need to find the length of the hypotenuse PR using the Pythagorean theorem.

According to the Pythagorean theorem:

$PR^2 = PQ^2 + QR^2$

Substitute the given values of PQ and QR:

$PR^2 = (4)^2 + (3)^2$

$PR^2 = 16 + 9$

$PR^2 = 25$

Taking the square root of both sides:

$PR = \sqrt{25}$

$PR = 5$ cm

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Now, we can find the value of $\sin R$. The sine of an acute angle in a right-angled triangle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse.

$\sin R = \frac{\text{Side opposite to angle R}}{\text{Hypotenuse}}$

In triangle PQR, the side opposite to angle R is PQ, and the hypotenuse is PR.

$\sin R = \frac{PQ}{PR}$

Substitute the values PQ = 4 cm and PR = 5 cm:

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

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Thus, the value of $\sin R$ is $\frac{4}{5}$.

Comparing this with the given options, we find that option (B) is correct.

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The correct option is (B) $\frac{4}{5}$.

Given:

$\tan A = \frac{8}{15}$

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To Find:

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

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Solution:

We are given $\tan A = \frac{8}{15}$. In a right-angled triangle, $\tan A$ is defined as the ratio of the side opposite to angle A to the side adjacent to angle A.

Let the side opposite to angle A be $8k$ and the side adjacent to angle A be $15k$, where $k$ is a positive constant.

Using the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let the hypotenuse be $h$.

$h^2 = (\text{Opposite side})^2 + (\text{Adjacent side})^2$

$h^2 = (8k)^2 + (15k)^2$

$h^2 = 64k^2 + 225k^2$

$h^2 = 289k^2$

Taking the square root of both sides:

$h = \sqrt{289k^2}$

$h = 17k$

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Now, we need to find the value of $\text{cosec } A$. The cosecant of angle A is defined as the reciprocal of the sine of angle A.

$\text{cosec } A = \frac{1}{\sin A}$

The sine of angle A is defined as the ratio of the side opposite to angle A to the hypotenuse.

$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{8k}{17k} = \frac{8}{17}$

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Now, substitute the value of $\sin A$ into the formula for $\text{cosec } A$:

$\text{cosec } A = \frac{1}{\sin A} = \frac{1}{8/17} = \frac{17}{8}$

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Thus, the value of $\text{cosec } A$ is $\frac{17}{8}$.

Comparing this with the given options, we find that option (C) is correct.

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The correct option is (C) $\frac{17}{8}$.

Given expression:

$\sin 30^\circ + \cos 60^\circ$

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Solution:

We know the standard trigonometric values for the angles $30^\circ$ and $60^\circ$.

The value of $\sin 30^\circ$ is $\frac{1}{2}$.

The value of $\cos 60^\circ$ is $\frac{1}{2}$.

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Substitute these values into the given expression:

$\sin 30^\circ + \cos 60^\circ = \frac{1}{2} + \frac{1}{2}$

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Now, perform the addition:

$\frac{1}{2} + \frac{1}{2} = \frac{1+1}{2} = \frac{2}{2} = 1$

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So, the value of $\sin 30^\circ + \cos 60^\circ$ is 1.

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Comparing this result with the given options, we find that option (B) is correct.

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The correct option is (B) 1.

Given expression:

$\frac{\tan 45^\circ}{\sin 90^\circ}$

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Solution:

We need to find the values of $\tan 45^\circ$ and $\sin 90^\circ$.

The standard trigonometric value for $\tan 45^\circ$ is 1.

$\tan 45^\circ = 1$

The standard trigonometric value for $\sin 90^\circ$ is 1.

$\sin 90^\circ = 1$

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Now, substitute these values into the given expression:

$\frac{\tan 45^\circ}{\sin 90^\circ} = \frac{1}{1}$

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Calculate the value of the fraction:

$\frac{1}{1} = 1$

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So, the value of $\frac{\tan 45^\circ}{\sin 90^\circ}$ is 1.

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Comparing this result with the given options, we find that option (B) is correct.

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The correct option is (B) 1.

Given:

An acute angle $\theta$.

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To Find:

The value of $\sin (90^\circ - \theta)$.

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Solution:

This question involves trigonometric ratios of complementary angles.

For any acute angle $\theta$, the angle $(90^\circ - \theta)$ is its complementary angle.

There are standard identities relating the trigonometric ratios of an angle to the trigonometric ratios of its complementary angle.

One of these identities states that the sine of an angle is equal to the cosine of its complementary angle.

That is,

$\sin (90^\circ - \theta) = \cos \theta$

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Similarly, other identities for complementary angles include:

$\cos (90^\circ - \theta) = \sin \theta$

$\tan (90^\circ - \theta) = \cot \theta$

$\cot (90^\circ - \theta) = \tan \theta$

$\text{sec } (90^\circ - \theta) = \text{cosec } \theta$

$\text{cosec } (90^\circ - \theta) = \text{sec } \theta$

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From the identities, we directly see that $\sin (90^\circ - \theta) = \cos \theta$.

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Comparing this result with the given options, we find that option (C) is correct.

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The correct option is (C) $\cos \theta$.

Given expression:

$\sin^2 \theta + \cos^2 \theta$

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Solution:

The expression $\sin^2 \theta + \cos^2 \theta$ is one of the fundamental trigonometric identities.

This identity states that for any real angle $\theta$, the sum of the square of the sine of the angle and the square of the cosine of the angle is always equal to 1.

The identity is:

$\sin^2 \theta + \cos^2 \theta = 1$

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This identity holds true regardless of the value of $\theta$. It is derived directly from the Pythagorean theorem applied to a right-angled triangle or from the definition of sine and cosine on the unit circle.

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Therefore, the value of $\sin^2 \theta + \cos^2 \theta$ for any angle $\theta$ is 1.

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Comparing this result with the given options, we find that option (B) is correct.

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The correct option is (B) 1.

Given expression:

$\sec^2 \theta - \tan^2 \theta$

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Solution:

This expression relates the secant and tangent of an angle $\theta$. It is one of the fundamental trigonometric identities derived from the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$.

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We start with the identity:

$\sin^2 \theta + \cos^2 \theta = 1$

Divide all terms by $\cos^2 \theta$ (assuming $\cos \theta \neq 0$):

$\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}$

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Using the definitions $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$, we can rewrite the equation:

$(\frac{\sin \theta}{\cos \theta})^2 + 1 = (\frac{1}{\cos \theta})^2$

$\tan^2 \theta + 1 = \sec^2 \theta$

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Rearranging this identity to match the given expression, we subtract $\tan^2 \theta$ from both sides:

$\sec^2 \theta - \tan^2 \theta = 1$

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This identity holds true for any angle $\theta$ for which $\sec \theta$ and $\tan \theta$ are defined (i.e., $\cos \theta \neq 0$).

Thus, the value of $\sec^2 \theta - \tan^2 \theta$ is always 1.

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Comparing this result with the given options, we find that option (B) is correct.

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The correct option is (B) 1.

Given:

$\sin A = \frac{3}{5}$, where A is an acute angle.

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To Find:

The value of $\cos A$.

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Solution:

We can use the fundamental trigonometric identity that relates sine and cosine:

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

Substitute the given value of $\sin A$ into the identity:

$(\frac{3}{5})^2 + \cos^2 A = 1$

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

To find $\cos^2 A$, subtract $\frac{9}{25}$ from both sides of the equation:

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

Find a common denominator (25) for the terms on the right side:

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

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

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

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

$\cos A = \pm\sqrt{\frac{16}{25}}$

$\cos A = \pm\frac{4}{5}$

We are given that A is an acute angle. Acute angles are in the first quadrant ($0^\circ < A < 90^\circ$), where both sine and cosine functions are positive.

Therefore, we must choose the positive value for $\cos A$.

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

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Alternate Solution (Using a right-angled triangle):

Given $\sin A = \frac{3}{5}$. In a right-angled triangle, $\sin A$ is defined as the ratio of the length of the side opposite to angle A to the length of the hypotenuse.

So, let the length of the side opposite to angle A be $3k$ and the length of the hypotenuse be $5k$, where $k$ is a positive constant representing the scaling factor.

Let the side adjacent to angle A be $x$. By the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides:

$(\text{Adjacent})^2 + (\text{Opposite})^2 = (\text{Hypotenuse})^2$

$x^2 + (3k)^2 = (5k)^2$

$x^2 + 9k^2 = 25k^2$

Subtract $9k^2$ from both sides:

$x^2 = 25k^2 - 9k^2$

$x^2 = 16k^2$

Taking the square root of both sides (and since $x$ is a length, it must be positive):

$x = \sqrt{16k^2} = 4k$

The length of the side adjacent to angle A is $4k$.

Now, $\cos A$ is defined as the ratio of the length of the side adjacent to angle A to the length of the hypotenuse:

$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4k}{5k} = \frac{4}{5}$

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Both methods confirm that the value of $\cos A$ is $\frac{4}{5}$.

Comparing this result with the given options, we find that option (A) is correct.

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The correct option is (A) $\frac{4}{5}$.

To determine which trigonometric ratios are defined for $\theta = 0^\circ$, we need to consider the definitions and values of these ratios at $0^\circ$.

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1. Sine of $\theta$:

$\sin 0^\circ = 0$

The sine of $0^\circ$ is 0, which is a defined value.

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2. Cosine of $\theta$:

$\cos 0^\circ = 1$

The cosine of $0^\circ$ is 1, which is a defined value.

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3. Tangent of $\theta$:

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\tan 0^\circ = \frac{\sin 0^\circ}{\cos 0^\circ} = \frac{0}{1} = 0$

The tangent of $0^\circ$ is 0, which is a defined value.

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4. Cosecant of $\theta$:

$\text{cosec } \theta = \frac{1}{\sin \theta}$

$\text{cosec } 0^\circ = \frac{1}{\sin 0^\circ} = \frac{1}{0}$

Division by zero is undefined. Therefore, $\text{cosec } 0^\circ$ is undefined.

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5. Cotangent of $\theta$:

$\cot \theta = \frac{\cos \theta}{\sin \theta}$

$\cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0}$

Division by zero is undefined. Therefore, $\cot 0^\circ$ is undefined.

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Based on these values, the trigonometric ratios defined for $\theta = 0^\circ$ are sine, cosine, and tangent.

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From the given options, the defined ratios are (A) $\sin \theta$, (B) $\cos \theta$, and (C) $\tan \theta$.

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The correct options are (A) $\sin \theta$, (B) $\cos \theta$, and (C) $\tan \theta$.

Analysis of Assertion (A):

The assertion states that for any acute angle $\theta$, $\cos (90^\circ - \theta) = \sin \theta$. This is a fundamental trigonometric identity related to complementary angles. In a right-angled triangle, if one acute angle is $\theta$, the other acute angle is $90^\circ - \theta$. The sine of $\theta$ is defined as the ratio of the opposite side to the hypotenuse, and the cosine of $(90^\circ - \theta)$ is defined as the ratio of the adjacent side to the hypotenuse with respect to the angle $(90^\circ - \theta)$. The side opposite to $\theta$ is adjacent to $(90^\circ - \theta)$, and vice versa. Therefore, the ratio for $\sin \theta$ is the same as the ratio for $\cos (90^\circ - \theta)$. Thus, Assertion (A) is true.

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Analysis of Reason (R):

The reason states that sine and cosine are complementary ratios. Trigonometric ratios are called complementary if the ratio of an angle is equal to the co-ratio of its complementary angle. The pair (sine, cosine) exhibits this property, as do (tangent, cotangent) and (secant, cosecant). The relationship $\sin \theta = \cos (90^\circ - \theta)$ and $\cos \theta = \sin (90^\circ - \theta)$ demonstrates that sine and cosine are indeed complementary ratios. Thus, Reason (R) is true.

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Evaluating if R is the correct explanation for A:

The identity $\cos (90^\circ - \theta) = \sin \theta$ is a direct consequence of the definition of sine and cosine as complementary ratios. The fact that sine and cosine are complementary ratios means precisely that the cosine of an angle equals the sine of its complement, and the sine of an angle equals the cosine of its complement. Therefore, Reason (R) correctly explains why Assertion (A) is true.

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Conclusion:

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

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

Analysis of Assertion (A):

The assertion is $\tan 45^\circ = 1$. This is a standard value in trigonometry derived from the properties of a $45^\circ-45^\circ-90^\circ$ right triangle or from the unit circle definition. The value of $\tan 45^\circ$ is indeed 1.

Thus, Assertion (A) is true.

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Analysis of Reason (R):

The reason describes a right triangle with angles $45^\circ, 45^\circ, 90^\circ$. This type of triangle is an isosceles right triangle. In an isosceles triangle, the sides opposite the equal angles are equal in length. Since the two acute angles are $45^\circ$, the sides opposite these angles must be equal.

Thus, Reason (R) is true.

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Evaluating if R is the correct explanation for A:

The tangent of an acute angle in a right triangle is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.

In a $45^\circ-45^\circ-90^\circ$ triangle, let the length of the sides opposite the $45^\circ$ angles be $s$. As stated in the reason, these sides are equal.

For one of the $45^\circ$ angles:

Side opposite the angle = $s$

Side adjacent to the angle = $s$

Using the definition of tangent:

$\tan 45^\circ = \frac{\text{Side opposite}}{\text{Side adjacent}} = \frac{s}{s}$

Since $s$ is a length in a triangle, $s \neq 0$. Therefore,

$\tan 45^\circ = 1$

The fact that the opposite and adjacent sides are equal in a $45^\circ$ right triangle, as stated in Reason (R), is the direct reason why the ratio (tangent) is 1. Therefore, Reason (R) is the correct explanation for Assertion (A).

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Conclusion:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

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

In a right-angled triangle, for an acute angle A:

The side opposite to angle A is the Perpendicular.

The side adjacent to angle A (not the hypotenuse) is the Base.

The side opposite the right angle is the Hypotenuse.

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The definitions of the given trigonometric ratios are:

(i) $\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}}$. This matches option (c).

(ii) $\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$. This matches option (d).

(iii) $\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}}$. This matches option (e).

(iv) $\text{cosec } A$ is the reciprocal of $\sin A$. $\text{cosec } A = \frac{1}{\sin A} = \frac{1}{\frac{\text{Opposite}}{\text{Hypotenuse}}} = \frac{\text{Hypotenuse}}{\text{Opposite}}$. This matches option (b).

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Based on these matches, we have the pairings:

(i) - (c)

(ii) - (d)

(iii) - (e)

(iv) - (b)

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Comparing these pairings with the given multiple-choice options:

(A) (i)-(c), (ii)-(d), (iii)-(e), (iv)-(b) - Matches our pairings.

(B) (i)-(d), (ii)-(c), (iii)-(e), (iv)-(b) - Incorrect for (i) and (ii).

(C) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(e) - Incorrect for (iii) and (iv).

(D) (i)-(e), (ii)-(d), (iii)-(c), (iv)-(a) - Incorrect for (i), (iii), and (iv).

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The correct option is the one that lists the correct pairings.

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The correct option is (A) (i)-(c), (ii)-(d), (iii)-(e), (iv)-(b).

We need to find the numerical values for each trigonometric ratio given in Column A:

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(i) $\sin 30^\circ$

The standard value of $\sin 30^\circ$ is $\frac{1}{2}$.

This matches option (d) in Column B.

So, (i) - (d).

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(ii) $\cos 45^\circ$

The standard value of $\cos 45^\circ$ is $\frac{1}{\sqrt{2}}$.

This matches option (b) in Column B.

So, (ii) - (b).

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(iii) $\tan 60^\circ$

The standard value of $\tan 60^\circ$ is $\sqrt{3}$.

This matches option (a) in Column B.

So, (iii) - (a).

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(iv) $\sec 0^\circ$

The secant function is the reciprocal of the cosine function: $\sec \theta = \frac{1}{\cos \theta}$.

The standard value of $\cos 0^\circ$ is 1.

So, $\sec 0^\circ = \frac{1}{\cos 0^\circ} = \frac{1}{1} = 1$.

This matches option (c) in Column B.

So, (iv) - (c).

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Putting the matches together, we have:

(i) - (d)

(ii) - (b)

(iii) - (a)

(iv) - (c)

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

(A) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c) - This matches our result.

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

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

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

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The correct option is (A) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c).

Given:

The support beam, the horizontal ceiling, and the vertical wall form a right-angled triangle.

The angle the beam makes with the ceiling is $30^\circ$. This is an acute angle in the right triangle.

The distance from the wall to where the beam touches the ceiling is 5 metres. In the right triangle, this distance is the side adjacent to the $30^\circ$ angle (since the wall is vertical and the ceiling is horizontal, forming the right angle, the distance along the ceiling is adjacent to the angle at the ceiling).

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To Find:

The length of the support beam.

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Solution:

Let the right triangle be denoted as follows:

Let A be the angle the beam makes with the ceiling (given as $30^\circ$).

Let B be the angle at the wall (which is $90^\circ$).

Let C be the angle at the point where the beam touches the ceiling (which is A, $30^\circ$).

The side opposite to angle B (right angle) is the hypotenuse, which is the support beam. Let its length be $L$.

The side opposite to angle C (angle at the ceiling, $30^\circ$) is the vertical wall height.

The side adjacent to angle C ($30^\circ$) is the distance from the wall along the ceiling, which is given as 5 metres.

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We have the angle ($30^\circ$), the length of the adjacent side (5 m), and we want to find the length of the hypotenuse ($L$).

The trigonometric ratio that relates the adjacent side and the hypotenuse is the cosine function:

$\cos (\text{angle}) = \frac{\text{Length of Adjacent side}}{\text{Length of Hypotenuse}}$

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Substitute the given values into the formula:

$\cos 30^\circ = \frac{5 \text{ metres}}{L}$

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Now, we need to solve this equation for $L$. Multiply both sides by $L$:

$L \times \cos 30^\circ = 5$

Divide both sides by $\cos 30^\circ$ (since $30^\circ$ is an acute angle, $\cos 30^\circ \neq 0$):

$L = \frac{5}{\cos 30^\circ}$

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The length of the support beam is $\frac{5}{\cos 30^\circ}$ metres.

Comparing this expression with the given options:

(A) $5 \sin 30^\circ$ metres

(B) $5 / \cos 30^\circ$ metres

(C) $5 / \sin 30^\circ$ metres

(D) $5 \cos 30^\circ$ metres

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Our calculated expression matches option (B).

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The correct option is (B) $5 / \cos 30^\circ$ metres.

Given from Case Study (Question 15):

The setup forms a right-angled triangle with the ceiling and wall.

The angle the beam makes with the ceiling is $30^\circ$.

The distance from the wall to the point on the ceiling where the beam touches is 5 metres. This distance is the side adjacent to the $30^\circ$ angle.

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To Find:

The height of the wall where the beam touches it. This height represents the side opposite to the $30^\circ$ angle in the right triangle.

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Solution:

Let the angle the beam makes with the ceiling be $\theta = 30^\circ$.

Let the distance from the wall to the point on the ceiling be the adjacent side, denoted as Adjacent = 5 m.

Let the height of the wall be the opposite side, denoted as Opposite = $H$ m.

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In a right-angled triangle, the tangent of an acute angle is defined as the ratio of the length of the opposite side to the length of the adjacent side.

$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$

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Substitute the given values into the formula:

$\tan 30^\circ = \frac{H}{5}$

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To find the height $H$, multiply both sides of the equation by 5:

$H = 5 \times \tan 30^\circ$

$H = 5 \tan 30^\circ$ metres

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The height of the wall where the beam touches it is $5 \tan 30^\circ$ metres.

Comparing this result with the given options:

(A) $5 \tan 30^\circ$ metres

(B) $5 \cos 30^\circ$ metres

(C) $5 \sin 30^\circ$ metres

(D) $5 / \tan 30^\circ$ metres

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Our calculated expression matches option (A).

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The correct option is (A) $5 \tan 30^\circ$ metres.

Given:

Possible values for $\sin \theta$ are given as options.

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To Find:

Which of the given values is NOT a possible value for $\sin \theta$.

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Solution:

The sine function, $\sin \theta$, represents the ratio of the length of the side opposite an angle to the length of the hypotenuse in a right-angled triangle. Since the hypotenuse is always the longest side in a right triangle, the length of the opposite side can never be greater than the length of the hypotenuse.

Therefore, the ratio $\frac{\text{Opposite side}}{\text{Hypotenuse}}$ must always be between -1 and 1, inclusive, for any real angle $\theta$.

Mathematically, the range of the sine function is $[-1, 1]$.

This means that for any angle $\theta$, the value of $\sin \theta$ must satisfy the inequality:

$-1 \leq \sin \theta \leq 1$

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Now let's check each of the given options against this range:

(A) 0.5: The value 0.5 is within the range $[-1, 1]$ because $-1 \leq 0.5 \leq 1$. So, 0.5 is a possible value for $\sin \theta$ (e.g., $\sin 30^\circ = 0.5$).

(B) 1: The value 1 is within the range $[-1, 1]$ because $-1 \leq 1 \leq 1$. So, 1 is a possible value for $\sin \theta$ (e.g., $\sin 90^\circ = 1$).

(C) 1.5: The value 1.5 is NOT within the range $[-1, 1]$ because $1.5 > 1$. So, 1.5 is NOT a possible value for $\sin \theta$.

(D) 0: The value 0 is within the range $[-1, 1]$ because $-1 \leq 0 \leq 1$. So, 0 is a possible value for $\sin \theta$ (e.g., $\sin 0^\circ = 0$ or $\sin 180^\circ = 0$).

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The value that falls outside the possible range for $\sin \theta$ is 1.5.

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The correct option is (C) 1.5.

We need to examine each given equation to determine which one is not a fundamental trigonometric identity. A trigonometric identity is an equation that is true for all valid values of the variable(s).

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Let's analyze each option:

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

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

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

This is another Pythagorean trigonometric identity, valid for all values of $\theta$ where $\cos \theta \neq 0$. It can be derived from $\sin^2 \theta + \cos^2 \theta = 1$ by dividing by $\cos^2 \theta$.

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

This is the third Pythagorean trigonometric identity, valid for all values of $\theta$ where $\sin \theta \neq 0$. It can be derived from $\sin^2 \theta + \cos^2 \theta = 1$ by dividing by $\sin^2 \theta$.

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(D) $\sin \theta + \cos \theta = 1$

Let's test this equation with a specific value of $\theta$, say $\theta = 30^\circ$.

$\sin 30^\circ + \cos 30^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}$

Since $\sqrt{3} \approx 1.732$, $\frac{1 + \sqrt{3}}{2} \approx \frac{1 + 1.732}{2} = \frac{2.732}{2} \approx 1.366$.

Since $\frac{1 + \sqrt{3}}{2} \neq 1$, the equation $\sin \theta + \cos \theta = 1$ is not true for all values of $\theta$. Therefore, it is not a trigonometric identity.

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The first three options are well-known fundamental trigonometric identities, while the fourth option is not true for all angles $\theta$.

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The correct option is (D) $\sin \theta + \cos \theta = 1$.

Given expression:

$\text{cosec}^2 A - \cot^2 A$

---

Solution:

This expression is related to one of the fundamental Pythagorean trigonometric identities.

The three main Pythagorean identities are:

1. $\sin^2 \theta + \cos^2 \theta = 1$

2. $1 + \tan^2 \theta = \sec^2 \theta$

3. $1 + \cot^2 \theta = \text{cosec}^2 \theta$

---

The third identity, $1 + \cot^2 A = \text{cosec}^2 A$, directly involves $\cot^2 A$ and $\text{cosec}^2 A$.

We can rearrange this identity to isolate the difference between $\text{cosec}^2 A$ and $\cot^2 A$.

Subtract $\cot^2 A$ from both sides of the identity $1 + \cot^2 A = \text{cosec}^2 A$:

$1 = \text{cosec}^2 A - \cot^2 A$

Or, written in the order of the question:

$\text{cosec}^2 A - \cot^2 A = 1$

---

This identity holds true for any angle A for which $\text{cosec } A$ and $\cot A$ are defined (i.e., $\sin A \neq 0$).

Thus, the value of $\text{cosec}^2 A - \cot^2 A$ is always 1.

---

Comparing this result with the given options, we find that option (C) is correct.

---

The correct option is (C) 1.

Given:

An acute angle $A$.

---

To Find:

The value of $\sec (90^\circ - A)$.

---

Solution:

This question requires the use of trigonometric identities for complementary angles.

For any acute angle $\theta$, the trigonometric ratios of $(90^\circ - \theta)$ are related to the trigonometric ratios of $\theta$ by the following identities:

$\sin (90^\circ - \theta) = \cos \theta$

$\cos (90^\circ - \theta) = \sin \theta$

$\tan (90^\circ - \theta) = \cot \theta$

$\cot (90^\circ - \theta) = \tan \theta$

$\sec (90^\circ - \theta) = \text{cosec } \theta$

$\text{cosec } (90^\circ - \theta) = \sec \theta$

---

Applying the identity for the secant function with $\theta = A$, we get:

$\sec (90^\circ - A) = \text{cosec } A$

---

Thus, the value of $\sec (90^\circ - A)$ is $\text{cosec } A$ for an acute angle $A$.

Comparing this result with the given options, we find that option (D) is correct.

---

The correct option is (D) $\text{cosec } A$.

Given:

$\tan \theta = 1$

The range of $\theta$ is $0^\circ \le \theta \le 90^\circ$.

---

To Find:

The value of $\theta$ that satisfies the given condition within the specified range.

---

Solution:

We are looking for an angle $\theta$ in the first quadrant (or at the boundaries $0^\circ$ or $90^\circ$) such that its tangent is 1.

We need to recall the standard trigonometric values for common angles like $0^\circ, 30^\circ, 45^\circ, 60^\circ,$ and $90^\circ$.

---

Let's list the tangent values for these angles:

$\tan 0^\circ = 0$

$\tan 30^\circ = \frac{1}{\sqrt{3}}$

$\tan 45^\circ = 1$

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

$\tan 90^\circ$ is undefined.

---

Comparing the given condition $\tan \theta = 1$ with the standard values, we see that the tangent of $45^\circ$ is 1.

So, $\tan \theta = \tan 45^\circ$.

---

Since $45^\circ$ lies within the given range ($0^\circ \le 45^\circ \le 90^\circ$), the value of $\theta$ is $45^\circ$.

---

Comparing this result with the given options, we find that option (B) is correct.

---

The correct option is (B) $45^\circ$.

Given expression:

$(1-\sin^2 A) \sec^2 A$

---

Solution:

We can simplify this expression using fundamental trigonometric identities.

---

Recall the Pythagorean identity:

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

Rearranging this identity to solve for $1 - \sin^2 A$, we subtract $\sin^2 A$ from both sides:

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

---

Now, substitute this into the given expression:

$(1-\sin^2 A) \sec^2 A = (\cos^2 A) \sec^2 A$

---

Recall the reciprocal identity for $\sec A$:

$\sec A = \frac{1}{\cos A}$

Squaring both sides, we get:

$\sec^2 A = \frac{1}{\cos^2 A}$

---

Substitute this into the expression we have:

$(\cos^2 A) \sec^2 A = \cos^2 A \times \frac{1}{\cos^2 A}$

---

Assuming $\cos A \neq 0$ (so that $\sec A$ is defined and $\cos^2 A \neq 0$), we can cancel out the $\cos^2 A$ terms:

$\cos^2 A \times \frac{1}{\cos^2 A} = \cancel{\cos^2 A} \times \frac{1}{\cancel{\cos^2 A}} = 1$

---

The simplified value of the expression is 1.

---

Comparing this result with the given options, we find that option (D) is correct.

---

The correct option is (D) 1.

Given:

$\sin A = \cos B$, where A and B are acute angles ($0^\circ < A < 90^\circ$, $0^\circ < B < 90^\circ$).

---

To Find:

The value of $A + B$.

---

Solution:

We are given the relationship $\sin A = \cos B$.

---

We know the trigonometric identity for complementary angles which states that the cosine of an angle is equal to the sine of its complementary angle, and vice versa.

For any angle $\theta$, $\cos \theta = \sin (90^\circ - \theta)$ and $\sin \theta = \cos (90^\circ - \theta)$.

---

Using the identity $\cos B = \sin (90^\circ - B)$, we can rewrite the given equation:

$\sin A = \sin (90^\circ - B)$

---

Since A and B are acute angles ($0^\circ < A < 90^\circ$ and $0^\circ < B < 90^\circ$), the angles $A$ and $(90^\circ - B)$ are in the range $[0^\circ, 90^\circ)$. In this range, the sine function is injective (one-to-one), meaning if $\sin X = \sin Y$, then $X = Y$.

---

Therefore, from $\sin A = \sin (90^\circ - B)$, we can conclude that:

$A = 90^\circ - B$

---

Now, rearrange this equation to find the sum $A + B$. Add B to both sides of the equation:

$A + B = 90^\circ$

---

The sum of the angles A and B is $90^\circ$. This result is consistent with A and B being acute angles.

---

Comparing this result with the given options, we find that option (D) is correct.

---

The correct option is (D) $90^\circ$.

Given expression:

$(1+\tan^2 \theta)(1-\sin^2 \theta)$

---

Solution:

We can simplify the given expression using fundamental trigonometric identities.

---

Recall the Pythagorean identity relating sine and cosine:

$\sin^2 \theta + \cos^2 \theta = 1$

Rearranging this identity, we can write:

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

---

Recall another Pythagorean identity relating tangent and secant:

$1 + \tan^2 \theta = \sec^2 \theta$

---

Now, substitute these identities into the given expression:

$(1+\tan^2 \theta)(1-\sin^2 \theta) = (\sec^2 \theta)(\cos^2 \theta)$

---

Recall the reciprocal identity relating secant and cosine:

$\sec \theta = \frac{1}{\cos \theta}$

Squaring both sides:

$\sec^2 \theta = \frac{1}{\cos^2 \theta}$

---

Substitute this into the expression we obtained:

$(\sec^2 \theta)(\cos^2 \theta) = \left(\frac{1}{\cos^2 \theta}\right)(\cos^2 \theta)$

---

Assuming $\cos \theta \neq 0$ (which is required for $\tan \theta$ and $\sec \theta$ to be defined), we can multiply the terms:

$\frac{1}{\cos^2 \theta} \times \cos^2 \theta = \frac{\cancel{\cos^2 \theta}}{\cancel{\cos^2 \theta}} = 1$

---

Thus, the simplified value of the expression is 1.

---

Comparing this result with the given options, we find that option (D) is correct.

---

The correct option is (D) 1.

Given:

A tower stands vertically on the ground, forming a right angle with the ground.

The distance from the point on the ground to the foot of the tower is 15 m. This is the horizontal distance.

The angle of elevation of the top of the tower from this point is $60^\circ$.

---

To Find:

The height of the tower.

---

Solution:

Let's represent the situation with a right-angled triangle.

Let the height of the tower be $h$. This is the vertical side of the triangle, opposite to the angle of elevation.

Let the distance from the point on the ground to the foot of the tower be $d = 15$ m. This is the horizontal side of the triangle, adjacent to the angle of elevation.

The angle of elevation is $\theta = 60^\circ$.

---

We have the angle ($\theta$), the length of the adjacent side ($d$), and we want to find the length of the opposite side ($h$).

The trigonometric ratio that relates the opposite side and the adjacent side is the tangent function:

$\tan \theta = \frac{\text{Length of Opposite side}}{\text{Length of Adjacent side}}$

---

Substitute the given values into the formula:

$\tan 60^\circ = \frac{h}{15 \text{ m}}$

---

Now, we need to solve this equation for $h$. Multiply both sides by 15:

$h = 15 \times \tan 60^\circ$

---

We know the standard trigonometric value for $\tan 60^\circ$:

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

---

Substitute this value into the equation for $h$:

$h = 15 \times \sqrt{3}$

$h = 15 \sqrt{3}$ metres

---

The height of the tower is $15 \sqrt{3}$ metres.

Comparing this result with the given options, we find that option (A) is correct.

---

The correct option is (A) $15 \sqrt{3}$ m.

Given:

Height of the pole = 10 m

Angle the wire makes with the ground = $30^\circ$

---

To Find:

The length of the wire.

---

Solution:

Let the height of the pole be $h$ and the length of the wire be $L$. The pole stands vertically, so it forms a right angle with the ground. The wire, the pole, and the ground form a right-angled triangle.

In this triangle, the height of the pole ($h = 10$ m) is the side opposite to the angle the wire makes with the ground ($30^\circ$), and the length of the wire ($L$) is the hypotenuse.

---

We need a trigonometric ratio that relates the opposite side and the hypotenuse. This ratio is the sine function:

$\sin(\text{angle}) = \frac{\text{Length of Opposite side}}{\text{Length of Hypotenuse}}$

---

Substitute the given values into the formula:

$\sin 30^\circ = \frac{10 \text{ m}}{L}$

---

We know the standard trigonometric value of $\sin 30^\circ$:

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

---

Substitute this value into the equation:

$\frac{1}{2} = \frac{10}{L}$

---

To solve for $L$, we can cross-multiply:

$1 \times L = 2 \times 10$

$L = 20$

---

The length of the wire is 20 metres.

---

Comparing this result with the given options, we find that option (B) is correct.

---

The correct option is (B) 20 m.

Given expression:

$2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$

---

Solution:

To evaluate the given expression, we need to use the standard trigonometric values for the angles $45^\circ$, $30^\circ$, and $60^\circ$.

---

The required values are:

$\tan 45^\circ = 1$

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

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

---

Now, substitute these values into the expression:

$2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ = 2(\tan 45^\circ)^2 + (\cos 30^\circ)^2 - (\sin 60^\circ)^2$

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

---

Perform the squaring operations:

$(1)^2 = 1$

$\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{(\sqrt{3})^2}{(2)^2} = \frac{3}{4}$

---

Substitute the squared values back into the expression:

$= 2(1) + \frac{3}{4} - \frac{3}{4}$

$= 2 + \frac{3}{4} - \frac{3}{4}$

---

Perform the addition and subtraction:

Since $\frac{3}{4} - \frac{3}{4} = 0$, the expression simplifies to:

$= 2 + 0$

$= 2$

---

The value of the expression $2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$ is 2.

---

Comparing this result with the given options, we find that option (B) is correct.

---

The correct option is (B) 2.

Given:

$\sin A = \frac{1}{2}$, where A is an acute angle.

$\cos B = \frac{1}{2}$, where B is an acute angle.

---

To Find:

The value of $A+B$.

---

Solution:

We are given that $\sin A = \frac{1}{2}$. We need to find the acute angle A whose sine is $\frac{1}{2}$.

From the standard trigonometric values for acute angles, we know that:

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

Since A is an acute angle, there is only one such value.

Therefore, $A = 30^\circ$.

---

We are also given that $\cos B = \frac{1}{2}$. We need to find the acute angle B whose cosine is $\frac{1}{2}$.

From the standard trigonometric values for acute angles, we know that:

$\cos 60^\circ = \frac{1}{2}$

Since B is an acute angle, there is only one such value.

Therefore, $B = 60^\circ$.

---

Now, we need to find the sum $A+B$.

$A + B = 30^\circ + 60^\circ$

$A + B = 90^\circ$

---

The value of $A+B$ is $90^\circ$.

Comparing this result with the given options, we find that option (C) is correct.

---

The correct option is (C) $90^\circ$.

Given from Case Study:

The angle of elevation from the observer to the kite is $45^\circ$.

The horizontal distance from the observer to the point directly below the kite is 20 metres.

We are asked to ignore the observer's height, which means the observer is at ground level for this calculation.

---

To Find:

The height of the kite from the ground.

---

Solution:

The situation can be represented as a right-angled triangle. The height of the kite is the vertical side (opposite the angle of elevation), and the distance from the observer to the point directly below the kite is the horizontal side (adjacent to the angle of elevation).

Let the height of the kite be $h$. This is the opposite side.

Let the horizontal distance be $d = 20$ m. This is the adjacent side.

The angle of elevation is $\theta = 45^\circ$.

---

The trigonometric ratio that relates the opposite side and the adjacent side is the tangent function:

$\tan \theta = \frac{\text{Length of Opposite side}}{\text{Length of Adjacent side}}$

---

Substitute the given values into the formula:

$\tan 45^\circ = \frac{h}{20 \text{ m}}$

---

We know the standard trigonometric value for $\tan 45^\circ$:

$\tan 45^\circ = 1$

---

Substitute this value into the equation:

$1 = \frac{h}{20}$

---

To solve for $h$, multiply both sides by 20:

$h = 1 \times 20$

$h = 20$ metres

---

The height of the kite from the ground is 20 metres.

Comparing this result with the given options, we find that option (A) is correct.

---

The correct option is (A) 20 metres.

Given:

The trigonometric value $\cos 60^\circ$ and several options.

---

To Find:

Which of the given options are equal to $\cos 60^\circ$.

---

Solution:

First, let's determine the standard value of $\cos 60^\circ$.

$\cos 60^\circ = \frac{1}{2}$

---

Now, let's evaluate the value of each option:

---

(A) $\sin 30^\circ$:

The standard value of $\sin 30^\circ$ is $\frac{1}{2}$.

Since $\sin 30^\circ = \frac{1}{2}$ and $\cos 60^\circ = \frac{1}{2}$, Option (A) is equal to $\cos 60^\circ$.

---

(B) $\tan 45^\circ$:

The standard value of $\tan 45^\circ$ is 1.

Since $\tan 45^\circ = 1$ and $\cos 60^\circ = \frac{1}{2}$, Option (B) is NOT equal to $\cos 60^\circ$.

---

(C) $\cos (90^\circ - 30^\circ)$:

First, calculate the angle inside the cosine function: $90^\circ - 30^\circ = 60^\circ$.

So, the expression is $\cos 60^\circ$.

The value of $\cos 60^\circ$ is $\frac{1}{2}$.

Since $\cos (90^\circ - 30^\circ) = \cos 60^\circ = \frac{1}{2}$ and $\cos 60^\circ = \frac{1}{2}$, Option (C) is equal to $\cos 60^\circ$. This also demonstrates the complementary angle identity $\cos (90^\circ - \theta) = \sin \theta$, where $\theta = 30^\circ$, so $\cos (90^\circ - 30^\circ) = \sin 30^\circ$, which we already found to be $\frac{1}{2}$.

---

(D) $\frac{1}{2}$:

This is a numerical value that is directly equal to the value of $\cos 60^\circ$.

Since $\frac{1}{2} = \frac{1}{2}$, Option (D) is equal to $\cos 60^\circ$.

---

Based on the evaluation of each option, the values equal to $\cos 60^\circ$ are $\sin 30^\circ$, $\cos (90^\circ - 30^\circ)$, and $\frac{1}{2}$.

---

The correct options are (A) $\sin 30^\circ$, (C) $\cos (90^\circ - 30^\circ)$, and (D) $\frac{1}{2}$.

Given expression:

$\frac{1}{\sin \theta}$

---

Solution:

The trigonometric functions have reciprocal relationships. The reciprocal of sine ($\sin \theta$) is defined as cosecant ($\text{cosec } \theta$).

The definition is:

$\text{cosec } \theta = \frac{1}{\sin \theta}$, provided that $\sin \theta \neq 0$.

---

Similarly, the reciprocal of cosine ($\cos \theta$) is secant ($\sec \theta$), defined as $\sec \theta = \frac{1}{\cos \theta}$, and the reciprocal of tangent ($\tan \theta$) is cotangent ($\cot \theta$), defined as $\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$.

---

From the definition, the value of $\frac{1}{\sin \theta}$ is $\text{cosec } \theta$.

---

Comparing this result with the given options, we find that option (D) is correct.

---

The correct option is (D) $\text{cosec } \theta$.

Given expression:

$\frac{1}{1+\sin \theta} + \frac{1}{1-\sin \theta}$

---

Solution:

To simplify this expression, we need to combine the two fractions by finding a common denominator. The common denominator is the product of the two denominators, which is $(1+\sin \theta)(1-\sin \theta)$.

---

Combine the fractions:

$\frac{1}{1+\sin \theta} + \frac{1}{1-\sin \theta} = \frac{1 \times (1-\sin \theta)}{(1+\sin \theta)(1-\sin \theta)} + \frac{1 \times (1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$

$= \frac{1-\sin \theta}{(1+\sin \theta)(1-\sin \theta)} + \frac{1+\sin \theta}{(1-\sin \theta)(1+\sin \theta)}$

$= \frac{(1-\sin \theta) + (1+\sin \theta)}{(1+\sin \theta)(1-\sin \theta)}$

---

Simplify the numerator and the denominator:

Numerator: $(1-\sin \theta) + (1+\sin \theta) = 1 - \sin \theta + 1 + \sin \theta = 1 + 1 + (-\sin \theta + \sin \theta) = 2 + 0 = 2$

Denominator: $(1+\sin \theta)(1-\sin \theta)$. This is a difference of squares, $(a+b)(a-b) = a^2 - b^2$.

So, $(1+\sin \theta)(1-\sin \theta) = 1^2 - (\sin \theta)^2 = 1 - \sin^2 \theta$

---

Substitute the simplified numerator and denominator back into the fraction:

The expression becomes $\frac{2}{1 - \sin^2 \theta}$.

---

Recall the fundamental Pythagorean trigonometric identity:

$\sin^2 \theta + \cos^2 \theta = 1$

Rearranging this identity, we get:

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

---

Substitute $1 - \sin^2 \theta$ with $\cos^2 \theta$ in the expression:

$\frac{2}{1 - \sin^2 \theta} = \frac{2}{\cos^2 \theta}$

---

Recall the reciprocal identity for secant: $\sec \theta = \frac{1}{\cos \theta}$.

Squaring both sides: $\sec^2 \theta = \frac{1}{\cos^2 \theta}$.

---

So, $\frac{2}{\cos^2 \theta}$ can be written as $2 \times \frac{1}{\cos^2 \theta} = 2 \sec^2 \theta$.

---

Thus, the simplified expression is $2 \sec^2 \theta$.

---

Comparing this result with the given options, we find that option (A) is correct.

---

The correct option is (A) $2 \sec^2 \theta$.

Given:

$\sin \theta + \cos \theta = \sqrt{2}$

$\theta$ is an acute angle ($0^\circ < \theta < 90^\circ$).

---

To Find:

The value of $\theta$.

---

Solution:

We are given the equation $\sin \theta + \cos \theta = \sqrt{2}$. We need to find the acute angle $\theta$ that satisfies this equation.

Let's consider the standard trigonometric values for common acute angles:

For $\theta = 30^\circ$:

$\sin 30^\circ + \cos 30^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}$. This is not equal to $\sqrt{2}$.

---

For $\theta = 45^\circ$:

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

We can simplify $\frac{2}{\sqrt{2}}$ by multiplying the numerator and denominator by $\sqrt{2}$:

$\frac{2}{\sqrt{2}} = \frac{2 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

So, $\sin 45^\circ + \cos 45^\circ = \sqrt{2}$. This matches the given equation.

---

For $\theta = 60^\circ$:

$\sin 60^\circ + \cos 60^\circ = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3}+1}{2}$. This is not equal to $\sqrt{2}$.

---

For $\theta = 90^\circ$:

$\sin 90^\circ + \cos 90^\circ = 1 + 0 = 1$. This is not equal to $\sqrt{2}$.

---

The acute angle $\theta$ that satisfies the equation $\sin \theta + \cos \theta = \sqrt{2}$ is $45^\circ$.

---

Comparing this result with the given options, we find that option (B) is correct.

---

The correct option is (B) $45^\circ$.

To determine which trigonometric ratios are defined for $\theta = 90^\circ$, we need to consider the values of these ratios at $90^\circ$.

---

1. Sine of $\theta$:

$\sin 90^\circ = 1$

The sine of $90^\circ$ is 1, which is a defined value.

---

2. Cosine of $\theta$:

$\cos 90^\circ = 0$

The cosine of $90^\circ$ is 0, which is a defined value.

---

3. Tangent of $\theta$:

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\tan 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0}$

Division by zero is undefined. Therefore, $\tan 90^\circ$ is undefined.

---

4. Secant of $\theta$:

$\sec \theta = \frac{1}{\cos \theta}$

$\sec 90^\circ = \frac{1}{\cos 90^\circ} = \frac{1}{0}$

Division by zero is undefined. Therefore, $\sec 90^\circ$ is undefined.

---

5. Cotangent of $\theta$:

$\cot \theta = \frac{\cos \theta}{\sin \theta}$

$\cot 90^\circ = \frac{\cos 90^\circ}{\sin 90^\circ} = \frac{0}{1} = 0$

The cotangent of $90^\circ$ is 0, which is a defined value.

---

Based on these values, the trigonometric ratios defined for $\theta = 90^\circ$ are sine, cosine, and cotangent.

---

From the given options, the defined ratios are (A) $\sin \theta$, (B) $\cos \theta$, and (E) $\cot \theta$.

---

The correct options are (A) $\sin \theta$, (B) $\cos \theta$, and (E) $\cot \theta$.

Given options:

Angles: $0^\circ, 30^\circ, 45^\circ, 90^\circ$.

---

To Find:

The angle for which the tangent value is undefined.

---

Solution:

The tangent of an angle $\theta$ is defined as the ratio of the sine of the angle to the cosine of the angle:

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

---

The tangent value is undefined when the denominator, $\cos \theta$, is equal to zero.

We need to find which of the given angles has a cosine value of zero.

---

Let's check the cosine values for each option:

For $\theta = 0^\circ$, $\cos 0^\circ = 1$.

For $\theta = 30^\circ$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$.

For $\theta = 45^\circ$, $\cos 45^\circ = \frac{1}{\sqrt{2}}$.

For $\theta = 90^\circ$, $\cos 90^\circ = 0$.

---

The cosine value is zero for $\theta = 90^\circ$.

Therefore, the tangent of $90^\circ$ is $\tan 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0}$, which is undefined.

---

Thus, the angle which has its tangent value undefined among the given options is $90^\circ$.

---

Comparing this result with the given options, we find that option (D) is correct.

---

The correct option is (D) $90^\circ$.

Given:

Several numerical values.

---

To Find:

Which of the given values is NOT a valid value for $\sec \theta$ for any angle $\theta$ for which $\sec \theta$ is defined.

---

Solution:

The secant function, $\sec \theta$, is defined as the reciprocal of the cosine function, $\cos \theta$.

$\sec \theta = \frac{1}{\cos \theta}$

---

We know that for any real angle $\theta$, the value of $\cos \theta$ lies in the range $[-1, 1]$.

$-1 \leq \cos \theta \leq 1$

The secant function is defined when $\cos \theta \neq 0$.

---

Let's consider the possible values of $\frac{1}{\cos \theta}$ when $\cos \theta \neq 0$:

If $\cos \theta$ is positive, $0 < \cos \theta \leq 1$. Taking the reciprocal, we get $\frac{1}{\cos \theta} \geq \frac{1}{1}$, so $\sec \theta \geq 1$.

If $\cos \theta$ is negative, $-1 \leq \cos \theta < 0$. Taking the reciprocal, we get $\frac{1}{\cos \theta} \leq \frac{1}{-1}$, so $\sec \theta \leq -1$.

---

Thus, the range of the secant function is $(-\infty, -1] \cup [1, \infty)$. This means that the absolute value of $\sec \theta$ must be greater than or equal to 1:

$|\sec \theta| \geq 1$

---

Now, let's check each of the given options:

(A) 1: $|1| = 1$. Since $1 \geq 1$, this is a valid value for $\sec \theta$ (e.g., $\sec 0^\circ = 1$).

(B) 0.8: $|0.8| = 0.8$. Since $0.8 < 1$, this is NOT a valid value for $\sec \theta$. The range requires the absolute value to be 1 or greater.

(C) 2: $|2| = 2$. Since $2 \geq 1$, this is a valid value for $\sec \theta$ (e.g., $\sec 60^\circ = 2$).

(D) $\sqrt{2}$: $|\sqrt{2}| = \sqrt{2} \approx 1.414$. Since $1.414 \geq 1$, this is a valid value for $\sec \theta$ (e.g., $\sec 45^\circ = \sqrt{2}$).

---

The only value among the options that does not satisfy the condition $|\sec \theta| \geq 1$ is 0.8.

---

The correct option is (B) 0.8.

Given expression:

$\cos 0^\circ + \sin 90^\circ$

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Solution:

To evaluate the expression, we need to find the standard trigonometric values for $\cos 0^\circ$ and $\sin 90^\circ$.

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The standard value of $\cos 0^\circ$ is 1.

$\cos 0^\circ = 1$

The standard value of $\sin 90^\circ$ is 1.

$\sin 90^\circ = 1$

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Now, substitute these values into the given expression:

$\cos 0^\circ + \sin 90^\circ = 1 + 1$

$= 2$

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The value of the expression $\cos 0^\circ + \sin 90^\circ$ is 2.

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Comparing this result with the given options, we find that option (C) is correct.

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The correct option is (C) 2.

Given:

$\text{cosec } A = \sec B$, where A and B are acute angles ($0^\circ < A < 90^\circ$, $0^\circ < B < 90^\circ$).

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To Find:

The value of $A + B$.

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Solution:

We are given the relationship $\text{cosec } A = \sec B$.

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We can use the trigonometric identities for complementary angles.

One such identity states that the cosecant of an angle is equal to the secant of its complementary angle:

$\text{cosec } \theta = \sec (90^\circ - \theta)$

Using this identity with $\theta = A$, we have:

$\text{cosec } A = \sec (90^\circ - A)$

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Substitute this into the given equation $\text{cosec } A = \sec B$:

$\sec (90^\circ - A) = \sec B$

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Since A and B are acute angles, $(90^\circ - A)$ will also be an acute angle or $0^\circ$. Specifically, since $0^\circ < A < 90^\circ$, we have $-90^\circ < -A < 0^\circ$, and adding $90^\circ$, we get $0^\circ < 90^\circ - A < 90^\circ$. Similarly, $0^\circ < B < 90^\circ$.

In the range $(0^\circ, 90^\circ)$, the secant function is injective (one-to-one), meaning if $\sec X = \sec Y$, then $X = Y$.

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Therefore, from $\sec (90^\circ - A) = \sec B$, we can conclude that:

$90^\circ - A = B$

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Now, rearrange this equation to find the sum $A + B$. Add A to both sides of the equation:

$90^\circ = A + B$

So, $A + B = 90^\circ$.

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Alternatively, we could use the identity $\sec B = \text{cosec } (90^\circ - B)$. Substituting this into the given equation $\text{cosec } A = \sec B$:

$\text{cosec } A = \text{cosec } (90^\circ - B)$

Since A and $(90^\circ - B)$ are in the range $(0^\circ, 90^\circ)$, where the cosecant function is injective, we have:

$A = 90^\circ - B$

Adding B to both sides: $A + B = 90^\circ$.

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Both methods yield the same result. The sum of the angles A and B is $90^\circ$.

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Comparing this result with the given options, we find that option (D) is correct.

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The correct option is (D) $90^\circ$.

Given expression:

$\tan^2 60^\circ + 2 \tan^2 45^\circ$

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Solution:

To evaluate the given expression, we need to use the standard trigonometric values for the angles $60^\circ$ and $45^\circ$.

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The required values are:

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

$\tan 45^\circ = 1$

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Now, substitute these values into the expression:

$\tan^2 60^\circ + 2 \tan^2 45^\circ = (\tan 60^\circ)^2 + 2 (\tan 45^\circ)^2$

$= (\sqrt{3})^2 + 2 (1)^2$

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Perform the squaring operations:

$(\sqrt{3})^2 = 3$

$(1)^2 = 1$

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Substitute the squared values back into the expression:

$= 3 + 2 (1)$

$= 3 + 2$

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Perform the addition:

$= 5$

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The value of the expression $\tan^2 60^\circ + 2 \tan^2 45^\circ$ is 5.

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Comparing this result with the given options, we find that option (C) is correct.

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The correct option is (C) 5.

Given:

An acute angle $\theta$, which means $0^\circ < \theta < 90^\circ$.

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To Find:

Which of the given statements is TRUE for the given range of $\theta$.

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Solution:

Let's analyze the behavior of each trigonometric function as $\theta$ increases from $0^\circ$ to $90^\circ$. We can consider a right-angled triangle or the unit circle to understand how the values change.

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Consider the interval $0^\circ < \theta < 90^\circ$:

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Analysis of Statement (A): As $\theta$ increases, $\sin \theta$ increases.

Let's look at some standard values:

$\sin 0^\circ = 0$

$\sin 30^\circ = 0.5$

$\sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707$

$\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$

$\sin 90^\circ = 1$

As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\sin \theta$ increases from 0 to 1. This statement is TRUE.

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Analysis of Statement (B): As $\theta$ increases, $\cos \theta$ increases.

Let's look at some standard values:

$\cos 0^\circ = 1$

$\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$

$\cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707$

$\cos 60^\circ = 0.5$

$\cos 90^\circ = 0$

As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\cos \theta$ decreases from 1 to 0. This statement is FALSE.

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Analysis of Statement (C): As $\theta$ increases, $\tan \theta$ decreases.

Let's look at some standard values:

$\tan 0^\circ = 0$

$\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$

$\tan 45^\circ = 1$

$\tan 60^\circ = \sqrt{3} \approx 1.732$

As $\theta$ approaches $90^\circ$ from the left, $\tan \theta$ approaches infinity.

As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\tan \theta$ increases from 0 towards infinity. This statement is FALSE.

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Analysis of Statement (D): $\sin \theta = \cos \theta$ for all values of $\theta$ in this range.

We know that $\sin \theta = \cos \theta$ only when $\theta = 45^\circ$ in the range $0^\circ \leq \theta \leq 90^\circ$. It is not true for all values of $\theta$ in the given range $0^\circ < \theta < 90^\circ$ (e.g., $\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$, which are not equal). This statement is FALSE.

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Based on the analysis, only statement (A) is true for $0^\circ < \theta < 90^\circ$.

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The correct option is (A) As $\theta$ increases, $\sin \theta$ increases.

In a right triangle ABC, right-angled at B, consider angle A. The sides are defined as follows:

Hypotenuse: The side opposite the right angle (side AC).

Opposite side (to angle A): The side opposite to angle A (side BC).

Adjacent side (to angle A): The side adjacent to angle A that is not the hypotenuse (side AB).

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The definitions of $\sin A$, $\cos A$, and $\tan A$ in terms of the sides are:

$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{\text{BC}}{\text{AC}}$

$\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{\text{AB}}{\text{AC}}$

$\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{\text{BC}}{\text{AB}}$

Given:

In $\triangle PQR$, right-angled at Q,

PQ = 3 cm

QR = 4 cm

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To Find:

Length of PR, $\sin P$, and $\cos P$.

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Solution:

Since $\triangle PQR$ is a right-angled triangle at Q, we can use the Pythagorean theorem to find the length of the hypotenuse PR.

According to the Pythagorean theorem:

$(Hypotenuse)^2 = (Opposite\ side)^2 + (Adjacent\ side)^2$

In $\triangle PQR$, PR is the hypotenuse, PQ and QR are the legs.

$PR^2 = PQ^2 + QR^2$

Substitute the given values of PQ and QR:

$PR^2 = 9\ \text{cm}^2 + 16\ \text{cm}^2$

$PR^2 = 25\ \text{cm}^2$

Taking the square root of both sides:

$PR = \sqrt{25\ \text{cm}^2}$

$PR = 5\ \text{cm}$

So, the length of the hypotenuse PR is 5 cm.

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Now, we need to find $\sin P$ and $\cos P$. For angle P:

Opposite side = QR = 4 cm

Adjacent side = PQ = 3 cm

Hypotenuse = PR = 5 cm

Definition of $\sin P$:

$\sin P = \frac{\text{Opposite side}}{\text{Hypotenuse}}$

$\sin P = \frac{QR}{PR}$

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

Definition of $\cos P$:

$\cos P = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$

$\cos P = \frac{PQ}{PR}$

$\cos P = \frac{3}{5}$

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Result:

The length of PR is 5 cm.

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

$\cos P = \frac{3}{5}$

Given:

$\sin A = \frac{3}{4}$

A is an acute angle.

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To Find:

$\cos A$ and $\tan A$.

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Solution:

We are given that $\sin A = \frac{3}{4}$. In a right triangle, the sine of an angle is defined as the ratio of the length of the opposite side to the length of the hypotenuse.

$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{3}{4}$

Let us consider a right triangle, say $\triangle ABC$, right-angled at B, with angle A being the acute angle. Let the opposite side BC = 3k and the hypotenuse AC = 4k, where k is a positive constant. For simplicity, we can assume k=1, so BC = 3 and AC = 4.

Using the Pythagorean theorem in $\triangle ABC$:

$AB^2 + BC^2 = AC^2$

Substitute the values of BC and AC:

$AB^2 + 9 = 16$

$AB^2 = 16 - 9$

$AB^2 = 7$

$AB = \sqrt{7}$ (Since AB is a length, it must be positive)

Now we have the lengths of all sides of the triangle relative to angle A:

Opposite side (BC) = 3

Adjacent side (AB) = $\sqrt{7}$

Hypotenuse (AC) = 4

Now we can find $\cos A$ and $\tan A$ using their definitions:

$\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC}$

$\cos A = \frac{\sqrt{7}}{4}$

$\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB}$

$\tan A = \frac{3}{\sqrt{7}}$

To rationalize the denominator for $\tan A$:

$\tan A = \frac{3}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$

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Alternatively, using trigonometric identity:

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

Given $\sin A = \frac{3}{4}$.

$(\frac{3}{4})^2 + \cos^2 A = 1$

$\frac{9}{16} + \cos^2 A = 1$

$\cos^2 A = 1 - \frac{9}{16} = \frac{16-9}{16} = \frac{7}{16}$

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

$\cos A = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$

Now, we use the identity $\tan A = \frac{\sin A}{\cos A}$.

$\tan A = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{4} \times \frac{4}{\sqrt{7}} = \frac{3}{\sqrt{7}}$

Rationalizing the denominator:

$\tan A = \frac{3}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$

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Result:

$\cos A = \frac{\sqrt{7}}{4}$

$\tan A = \frac{3\sqrt{7}}{7}$

Given:

$15 \cot A = 8$

A is an acute angle.

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To Find:

$\sin A$ and $\sec A$.

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Solution (Using a right triangle):

From the given equation, we can find the value of $\cot A$:

$\cot A = \frac{8}{15}$

In a right triangle, $\cot A$ is defined as the ratio of the length of the adjacent side to the length of the opposite side for angle A.

$\cot A = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{8}{15}$

Consider a right triangle, say $\triangle ABC$, right-angled at B, with angle A being the acute angle. Let the adjacent side AB = 8k and the opposite side BC = 15k, where k is a positive constant. (For simplicity, we can assume k=1, so AB = 8 and BC = 15).

We need to find the length of the hypotenuse AC using the Pythagorean theorem:

$AC^2 = AB^2 + BC^2$

Substitute the values of AB and BC:

$AC^2 = 64k^2 + 225k^2$

$AC^2 = 289k^2$

Taking the square root of both sides (since AC is a length, it's positive):

$AC = \sqrt{289k^2} = 17k$

Now we have the lengths of all sides of the triangle relative to angle A:

Opposite side (BC) = 15k

Adjacent side (AB) = 8k

Hypotenuse (AC) = 17k

Now we can find $\sin A$ and $\sec A$ using their definitions:

$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC}$

$\sin A = \frac{15k}{17k} = \frac{15}{17}$

$\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent side}} = \frac{AC}{AB}$

$\sec A = \frac{17k}{8k} = \frac{17}{8}$

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Alternatively, using trigonometric identities:

We are given $\cot A = \frac{8}{15}$.

We know the identity $\text{cosec}^2 A = 1 + \cot^2 A$.

Substitute the value of $\cot A$:

$\text{cosec}^2 A = 1 + \frac{64}{225} = \frac{225 + 64}{225} = \frac{289}{225}$

Since A is an acute angle, $\text{cosec} A$ is positive.

$\text{cosec} A = \sqrt{\frac{289}{225}} = \frac{17}{15}$

We know that $\sin A = \frac{1}{\text{cosec} A}$.

$\sin A = \frac{1}{\frac{17}{15}} = \frac{15}{17}$

Now, we can find $\sec A$. We know the identity $\sec^2 A = 1 + \tan^2 A$.

First, find $\tan A$ from $\cot A$:

$\tan A = \frac{1}{\cot A} = \frac{1}{\frac{8}{15}} = \frac{15}{8}$

Substitute the value of $\tan A$ into the identity for $\sec A$:

$\sec^2 A = 1 + \frac{225}{64} = \frac{64 + 225}{64} = \frac{289}{64}$

Since A is an acute angle, $\sec A$ is positive.

$\sec A = \sqrt{\frac{289}{64}} = \frac{17}{8}$

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Result:

$\sin A = \frac{15}{17}$

$\sec A = \frac{17}{8}$

Given Expression:

$\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$

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Solution:

We need to substitute the standard values of the trigonometric ratios for $30^\circ$ and $60^\circ$.

The standard values are:

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

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

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

$\cos 60^\circ = \frac{1}{2}$

Substitute these values into the given expression:

Now, perform the multiplication:

Expression = $\frac{\sqrt{3} \times \sqrt{3}}{2 \times 2} + \frac{1 \times 1}{2 \times 2}$

Expression = $\frac{3}{4} + \frac{1}{4}$

Add the fractions:

Expression = $\frac{3+1}{4} = \frac{4}{4}$

Expression = $1$

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Alternatively (using trigonometric identity):

The given expression is in the form $\sin A \cos B + \cos A \sin B$, which is the expansion of $\sin(A+B)$.

Here, $A = 60^\circ$ and $B = 30^\circ$.

So, $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \sin(60^\circ + 30^\circ)$

= $\sin(90^\circ)$

The value of $\sin 90^\circ$ is 1.

Thus, the value of the expression is 1.

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Result:

The value of the expression $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$ is 1.

Given Expression:

$2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$

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Solution:

We need to substitute the standard values of the trigonometric ratios for $45^\circ$, $30^\circ$, and $60^\circ$.

The standard values are:

$\tan 45^\circ = 1$

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

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

Substitute these values into the given expression:

Now, evaluate the squares:

$(1)^2 = 1$

$\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{(\sqrt{3})^2}{(2)^2} = \frac{3}{4}$

Substitute the squared values back into the expression:

Expression = $2(1) + \frac{3}{4} - \frac{3}{4}$

Perform the operations:

Expression = $2 + \frac{3}{4} - \frac{3}{4}$

Expression = $2 + 0$

Expression = $2$

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Alternatively (not really an alternative method, just noting cancellation):

Notice that $\cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$ and $\sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$.

So, $\cos^2 30^\circ - \sin^2 60^\circ = \frac{3}{4} - \frac{3}{4} = 0$.

The expression becomes $2 \tan^2 45^\circ + 0$.

Since $\tan 45^\circ = 1$, $\tan^2 45^\circ = 1^2 = 1$.

Expression = $2(1) + 0 = 2$.

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Result:

The value of the expression $2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$ is 2.

Given Expression:

$\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$

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Solution:

We need to evaluate the given expression by substituting the standard value of $\tan 30^\circ$.

The standard value of $\tan 30^\circ$ is $\frac{1}{\sqrt{3}}$.

Substitute this value into the expression:

Numerator = $2 \tan 30^\circ = 2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$

Denominator = $1 + \tan^2 30^\circ = 1 + \left(\frac{1}{\sqrt{3}}\right)^2 = 1 + \frac{1}{3}$

Denominator = $\frac{3}{3} + \frac{1}{3} = \frac{3+1}{3} = \frac{4}{3}$

Now divide the numerator by the denominator:

Expression = $\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$

Expression = $\frac{2}{\sqrt{3}} \times \frac{3}{4}$

Expression = $\frac{2 \times 3}{\sqrt{3} \times 4}$

Expression = $\frac{6}{4\sqrt{3}}$

Simplify the fraction by cancelling common factors:

Expression = $\frac{\cancel{6}^{3}}{\cancel{4}_{2}\sqrt{3}} = \frac{3}{2\sqrt{3}}$

Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$:

Expression = $\frac{3}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{2 \times 3}$

Expression = $\frac{\cancel{3}\sqrt{3}}{2\cancel{3}} = \frac{\sqrt{3}}{2}$

The value of the expression is $\frac{\sqrt{3}}{2}$.

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Justification of Choice:

We compare this value with the standard trigonometric ratios:

$\sin 60^\circ = \frac{\sqrt{3}}{2}$

$\cos 60^\circ = \frac{1}{2}$

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

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

The value $\frac{\sqrt{3}}{2}$ matches $\sin 60^\circ$.

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Alternatively (using trigonometric identity):

The expression $\frac{2 \tan A}{1 + \tan^2 A}$ is a trigonometric identity that is equal to $\sin 2A$.

In the given expression, $A = 30^\circ$.

Substitute $A = 30^\circ$ into the identity:

Expression = $\sin 60^\circ$

The standard value of $\sin 60^\circ$ is $\frac{\sqrt{3}}{2}$.

So, the value of the expression is $\frac{\sqrt{3}}{2}$.

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Result:

The value of the expression is $\frac{\sqrt{3}}{2}$, which is equal to $\sin 60^\circ$.

The correct option is $\sin 60^\circ$.

Given Expression:

$\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$

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Solution:

We need to evaluate the given expression by substituting the standard values of the trigonometric ratios for $45^\circ$ and $30^\circ$.

The standard values are:

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

$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$

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

Substitute these values into the given expression:

First, simplify the denominator:

Denominator = $\frac{2}{\sqrt{3}} + 2 = \frac{2}{\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3}} = \frac{2 + 2\sqrt{3}}{\sqrt{3}}$

Now substitute the simplified denominator back into the expression:

Expression = $\frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}}$

Divide the fractions by multiplying the numerator by the reciprocal of the denominator:

Expression = $\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 + 2\sqrt{3}}$

Expression = $\frac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})}$

Expression = $\frac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}}$

Now, rationalize the denominator. We can factor out 2 from the denominator first:

Expression = $\frac{\sqrt{3}}{2(\sqrt{2} + \sqrt{6})}$

Multiply the numerator and the denominator by the conjugate of $(\sqrt{2} + \sqrt{6})$, which is $(\sqrt{2} - \sqrt{6})$:

Expression = $\frac{\sqrt{3}}{2(\sqrt{2} + \sqrt{6})} \times \frac{\sqrt{2} - \sqrt{6}}{\sqrt{2} - \sqrt{6}}$

Numerator = $\sqrt{3}(\sqrt{2} - \sqrt{6}) = \sqrt{3}\sqrt{2} - \sqrt{3}\sqrt{6} = \sqrt{6} - \sqrt{18}$

Simplify $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$.

Numerator = $\sqrt{6} - 3\sqrt{2}$

Denominator = $2(\sqrt{2} + \sqrt{6})(\sqrt{2} - \sqrt{6})$

Using the identity $(a+b)(a-b) = a^2 - b^2$:

Denominator = $2((\sqrt{2})^2 - (\sqrt{6})^2) = 2(2 - 6) = 2(-4) = -8$

Substitute the simplified numerator and denominator back into the expression:

Expression = $\frac{\sqrt{6} - 3\sqrt{2}}{-8}$

We can rewrite this by changing the sign of the numerator and the denominator:

Expression = $\frac{-(\sqrt{6} - 3\sqrt{2})}{8} = \frac{3\sqrt{2} - \sqrt{6}}{8}$

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Result:

The value of the expression $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$ is $\frac{3\sqrt{2} - \sqrt{6}}{8}$.

Given:

$\sin (A - B) = \frac{1}{2}$

$\cos (A + B) = \frac{1}{2}$

$0^\circ < A + B \leq 90^\circ$ and $A > B$.

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To Find:

The values of A and B.

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Solution:

We are given that $\sin (A - B) = \frac{1}{2}$. We know that $\sin 30^\circ = \frac{1}{2}$.

Therefore, since $A-B$ must correspond to this value within a relevant range (implied by the conditions on A and B), we can write:

We are also given that $\cos (A + B) = \frac{1}{2}$. We know that $\cos 60^\circ = \frac{1}{2}$.

Given the condition $0^\circ < A + B \leq 90^\circ$, we can write:

Now we have a system of two linear equations with two variables A and B:

Equation (1): $A - B = 30^\circ$

Equation (2): $A + B = 60^\circ$

We can solve this system by adding equation (1) and equation (2):

$(A - B) + (A + B) = 30^\circ + 60^\circ$

$A - B + A + B = 90^\circ$

$2A = 90^\circ$

$A = \frac{90^\circ}{2}$

$A = 45^\circ$

Now, substitute the value of A into equation (2) to find the value of B:

$B = 60^\circ - 45^\circ$

$B = 15^\circ$

Let's check if the obtained values satisfy the given conditions:

$A + B = 45^\circ + 15^\circ = 60^\circ$. This satisfies $0^\circ < A + B \leq 90^\circ$.

$A = 45^\circ$ and $B = 15^\circ$. This satisfies $A > B$.

The values satisfy all the given conditions.

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Result:

The values of A and B are $A = 45^\circ$ and $B = 15^\circ$.

Given Expression:

$\frac{\sin 18^\circ}{\cos 72^\circ}$

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To Evaluate:

The value of the given expression.

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Solution:

We notice that the angles in the numerator and the denominator are complementary, i.e., $18^\circ + 72^\circ = 90^\circ$.

We can use the trigonometric identity relating trigonometric ratios of complementary angles.

The identity is $\cos (90^\circ - \theta) = \sin \theta$ or $\sin (90^\circ - \theta) = \cos \theta$.

Let's express the denominator $\cos 72^\circ$ in terms of sine:

$\cos 72^\circ = \cos (90^\circ - 18^\circ)$

Using the identity $\cos (90^\circ - \theta) = \sin \theta$ with $\theta = 18^\circ$:

Now substitute this into the given expression:

Expression = $\frac{\sin 18^\circ}{\cos 72^\circ}$

Assuming $\sin 18^\circ \neq 0$ (which is true), we can cancel out the term in the numerator and denominator.

Expression = $\frac{\cancel{\sin 18^\circ}}{\cancel{\sin 18^\circ}} = 1$

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Alternatively:

We can express the numerator $\sin 18^\circ$ in terms of cosine:

$\sin 18^\circ = \sin (90^\circ - 72^\circ)$

Using the identity $\sin (90^\circ - \theta) = \cos \theta$ with $\theta = 72^\circ$:

Now substitute this into the given expression:

Expression = $\frac{\sin 18^\circ}{\cos 72^\circ}$

Assuming $\cos 72^\circ \neq 0$ (which is true), we can cancel out the term.

Expression = $\frac{\cancel{\cos 72^\circ}}{\cancel{\cos 72^\circ}} = 1$

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Result:

The value of the expression $\frac{\sin 18^\circ}{\cos 72^\circ}$ is 1.

Given Expression:

$\tan 26^\circ \cot 64^\circ$

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To Evaluate:

The value of the given expression.

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Solution:

We observe that the angles $26^\circ$ and $64^\circ$ are complementary, as their sum is $26^\circ + 64^\circ = 90^\circ$.

We can use the trigonometric identities relating trigonometric ratios of complementary angles.

Specifically, we use the identity $\cot (90^\circ - \theta) = \tan \theta$.

Let $\theta = 26^\circ$. Then $90^\circ - \theta = 90^\circ - 26^\circ = 64^\circ$.

Using the identity $\cot (90^\circ - \theta) = \tan \theta$, we have:

Now substitute the result from (i) into the given expression:

Expression = $\tan 26^\circ \times \cot 64^\circ$

Expression = $\tan^2 26^\circ$

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Alternatively:

We can use the identity $\tan (90^\circ - \theta) = \cot \theta$.

Let $\theta = 64^\circ$. Then $90^\circ - \theta = 90^\circ - 64^\circ = 26^\circ$.

Using the identity $\tan (90^\circ - \theta) = \cot \theta$, we have:

Substitute the result from (iii) into the given expression:

Expression = $\tan 26^\circ \times \cot 64^\circ$

Expression = $\cot^2 64^\circ$

Note that $\tan^2 26^\circ = (\tan 26^\circ)^2$ and $\cot^2 64^\circ = (\cot 64^\circ)^2$. Since $\tan 26^\circ = \cot 64^\circ$, these two results are equivalent.

A typical problem involving the product of tan and cot of complementary angles simplifies to 1 only if the product is $\tan A \times \tan B$ or $\cot A \times \cot B$ where $A+B=90^\circ$, because $\tan A \tan (90-A) = \tan A \cot A = 1$. In this case, the expression is $\tan A \cot B$, where $A+B=90$. Since $\cot B = \tan A$, the product is $\tan A \times \tan A = \tan^2 A$.

---

Result:

The value of the expression $\tan 26^\circ \cot 64^\circ$ is $\tan^2 26^\circ$ (or equivalently, $\cot^2 64^\circ$).

Given:

$\sin 3A = \cos (A - 26^\circ)$

3A is an acute angle, which means $0^\circ < 3A < 90^\circ$.

---

To Find:

The value of A.

---

Solution:

We are given the equation $\sin 3A = \cos (A - 26^\circ)$.

We use the trigonometric identity for complementary angles: $\sin \theta = \cos (90^\circ - \theta)$.

Using this identity, we can rewrite the left side of the equation:

Now, substitute this into the given equation:

From $\sin 3A = \cos (A - 26^\circ)$ and equation (i), we get:

Since 3A is an acute angle ($0^\circ < 3A < 90^\circ$), it follows that $90^\circ - 3A$ is also an acute angle ($0^\circ < 90^\circ - 3A < 90^\circ$).

For angles in the range $[0^\circ, 180^\circ]$, if $\cos X = \cos Y$, then $X=Y$. In this case, since $90^\circ - 3A$ is acute and its cosine is positive, $A - 26^\circ$ must be an angle whose cosine is equal to $\cos (90^\circ - 3A)$. The simplest solution arises when $A - 26^\circ$ equals $90^\circ - 3A$ (or differs by a multiple of $360^\circ$, which is not relevant here for finding the value of A in the typical context).

Equating the angles from equation (ii):

Now, we solve this linear equation for A:

Add $3A$ to both sides and add $26^\circ$ to both sides:

$90^\circ + 26^\circ = A + 3A$

$116^\circ = 4A$

Divide both sides by 4:

$A = \frac{116^\circ}{4}$

$A = 29^\circ$

Let's verify if the condition that 3A is an acute angle is satisfied:

$3A = 3 \times 29^\circ = 87^\circ$.

Since $0^\circ < 87^\circ < 90^\circ$, 3A is indeed an acute angle. The value of A is consistent with the given condition.

---

Result:

The value of A is $29^\circ$.

Given:

$\tan A = \cot B$

(Assuming A and B are acute angles, which is typical in this context).

---

To Prove:

$A + B = 90^\circ$

---

Proof:

We are given that $\tan A = \cot B$.

We know the trigonometric identity for complementary angles:

Using identity (i), we can rewrite $\cot B$ as $\tan (90^\circ - B)$.

Now, substitute the result from equation (ii) into the given equation $\tan A = \cot B$:

If the tangents of two angles are equal, and the angles are acute, then the angles must be equal.

From equation (iii), we can equate the angles:

Rearrange equation (iv) by adding B to both sides:

A + B = $90^\circ$

Thus, we have proven that if $\tan A = \cot B$, then $A + B = 90^\circ$ (under the assumption that A and B are acute angles).

---

Result:

It is proven that $A + B = 90^\circ$ when $\tan A = \cot B$.

Given Expression:

$\sin 67^\circ + \cos 75^\circ$

---

To Express:

The expression in terms of trigonometric ratios of angles between $0^\circ$ and $45^\circ$.

---

Solution:

We need to change the angles $67^\circ$ and $75^\circ$ into their complementary angles, which will fall between $0^\circ$ and $45^\circ$. We use the complementary angle identities:

Consider the first term, $\sin 67^\circ$. Here $\theta = 67^\circ$. Using identity (i):

$\sin 67^\circ = \cos 23^\circ$

The angle $23^\circ$ is between $0^\circ$ and $45^\circ$.

Consider the second term, $\cos 75^\circ$. Here $\theta = 75^\circ$. Using identity (ii):

$\cos 75^\circ = \sin 15^\circ$

The angle $15^\circ$ is between $0^\circ$ and $45^\circ$.

Now, substitute the results from (iii) and (iv) back into the original expression:

$\sin 67^\circ + \cos 75^\circ = (\cos 23^\circ) + (\sin 15^\circ)$

So, $\sin 67^\circ + \cos 75^\circ = \cos 23^\circ + \sin 15^\circ$.

The angles in the resulting expression are $23^\circ$ and $15^\circ$, which are both between $0^\circ$ and $45^\circ$.

---

Result:

The expression $\sin 67^\circ + \cos 75^\circ$ can be expressed as $\cos 23^\circ + \sin 15^\circ$, where the angles are between $0^\circ$ and $45^\circ$.

Given:

An identity to prove: $\sin^2 \theta + \cos^2 \theta = 1$.

We will prove this using a right-angled triangle for an acute angle $\theta$. The identity holds true for all real values of $\theta$.

---

To Prove:

$\sin^2 \theta + \cos^2 \theta = 1$

---

Proof:

Consider a right-angled triangle ABC, right-angled at B. Let $\angle A = \theta$, where $\theta$ is an acute angle.

In $\triangle ABC$:

The side opposite to angle $\theta$ is BC.

The side adjacent to angle $\theta$ is AB.

The hypotenuse is AC.

By the definition of trigonometric ratios:

According to the Pythagorean theorem in the right-angled triangle ABC:

Now, consider the Left Hand Side (LHS) of the identity: $\sin^2 \theta + \cos^2 \theta$.

Substitute the expressions for $\sin \theta$ and $\cos \theta$ from equations (1) and (2) into the LHS:

Simplify the expression:

$\text{LHS} = \frac{(BC)^2}{(AC)^2} + \frac{(AB)^2}{(AC)^2}$

$\text{LHS} = \frac{(BC)^2 + (AB)^2}{(AC)^2}$

From the Pythagorean theorem (equation 3), we know that $(AB)^2 + (BC)^2 = (AC)^2$. Substitute this into the numerator of the expression for LHS:

Assuming $AC \neq 0$ (which is true for a triangle), we can cancel out $(AC)^2$ from the numerator and denominator:

$\text{LHS} = 1$

The Right Hand Side (RHS) of the identity is also 1.

Since LHS = RHS, the identity $\sin^2 \theta + \cos^2 \theta = 1$ is proven for an acute angle $\theta$. This identity is fundamental and holds for all real numbers $\theta$ beyond the context of a right triangle.

---

Result:

The identity $\sin^2 \theta + \cos^2 \theta = 1$ is proved.

Given:

An identity to prove: $\sec^2 A - \tan^2 A = 1$.

(Assuming A is an angle for which $\sec A$ and $\tan A$ are defined, i.e., $\cos A \neq 0$).

---

To Prove:

$\sec^2 A - \tan^2 A = 1$

---

Proof:

We start with the fundamental trigonometric identity, which we have already proven:

Assuming $\cos A \neq 0$, we can divide both sides of equation (1) by $\cos^2 A$.

We use the definitions of $\tan A$ and $\sec A$:

$\tan A = \frac{\sin A}{\cos A}$, so $\tan^2 A = \left(\frac{\sin A}{\cos A}\right)^2 = \frac{\sin^2 A}{\cos^2 A}$.

$\sec A = \frac{1}{\cos A}$, so $\sec^2 A = \left(\frac{1}{\cos A}\right)^2 = \frac{1}{\cos^2 A}$.

Also, $\frac{\cos^2 A}{\cos^2 A} = 1$ (assuming $\cos A \neq 0$).

Substitute these into equation (2):

Rearrange equation (3) to match the required identity by subtracting $\tan^2 A$ from both sides:

$1 = \sec^2 A - \tan^2 A$

Or, writing it in the standard form:

$\sec^2 A - \tan^2 A = 1$

Thus, the identity is proven.

---

Result:

The identity $\sec^2 A - \tan^2 A = 1$ is proved.

Given:

An identity to prove: $\text{cosec}^2 A - \cot^2 A = 1$.

(Assuming A is an angle for which $\text{cosec } A$ and $\cot A$ are defined, i.e., $\sin A \neq 0$).

---

To Prove:

$\text{cosec}^2 A - \cot^2 A = 1$

---

Proof:

We start with the fundamental trigonometric identity:

Assuming $\sin A \neq 0$, we can divide both sides of equation (1) by $\sin^2 A$.

We use the definitions of $\cot A$ and $\text{cosec } A$:

$\cot A = \frac{\cos A}{\sin A}$, so $\cot^2 A = \left(\frac{\cos A}{\sin A}\right)^2 = \frac{\cos^2 A}{\sin^2 A}$.

$\text{cosec } A = \frac{1}{\sin A}$, so $\text{cosec}^2 A = \left(\frac{1}{\sin A}\right)^2 = \frac{1}{\sin^2 A}$.

Also, $\frac{\sin^2 A}{\sin^2 A} = 1$ (assuming $\sin A \neq 0$).

Substitute these into equation (2):

Rearrange equation (3) to match the required identity by subtracting $\cot^2 A$ from both sides:

$1 = \text{cosec}^2 A - \cot^2 A$

Or, writing it in the standard form:

$\text{cosec}^2 A - \cot^2 A = 1$

Thus, the identity is proven.

---

Result:

The identity $\text{cosec}^2 A - \cot^2 A = 1$ is proved.

Given:

$\sec A = \frac{13}{12}$

A is an acute angle.

---

To Find:

The values of $\sin A$, $\cos A$, $\tan A$, $\cot A$, and $\text{cosec } A$.

---

Solution (Using a right triangle):

We are given that $\sec A = \frac{13}{12}$. In a right triangle, the secant of an angle is defined as the ratio of the length of the hypotenuse to the length of the adjacent side.

$\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent side}} = \frac{13}{12}$

Consider a right triangle, say $\triangle ABC$, right-angled at B, with angle A being the acute angle. Let the hypotenuse AC = 13k and the adjacent side AB = 12k, where k is a positive constant. (For simplicity in calculation, we can assume k=1, so AC = 13 and AB = 12).

We need to find the length of the opposite side BC using the Pythagorean theorem:

$AB^2 + BC^2 = AC^2$

Substitute the values of AB and AC:

$144k^2 + BC^2 = 169k^2$

$BC^2 = 169k^2 - 144k^2$

$BC^2 = 25k^2$

Taking the square root of both sides (since BC is a length, it must be positive):

$BC = \sqrt{25k^2} = 5k$

Now we have the lengths of all sides of the triangle relative to angle A:

Opposite side (BC) = 5k

Adjacent side (AB) = 12k

Hypotenuse (AC) = 13k

Now we can find the other trigonometric ratios using their definitions:

$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}$

$\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}$

$\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}$

$\cot A = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{AB}{BC} = \frac{12k}{5k} = \frac{12}{5}$

$\text{cosec } A = \frac{\text{Hypotenuse}}{\text{Opposite side}} = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5}$

---

Alternative Solution (Using trigonometric identities):

We are given $\sec A = \frac{13}{12}$.

1. Find $\cos A$ using the reciprocal identity $\cos A = \frac{1}{\sec A}$:

$\cos A = \frac{1}{\frac{13}{12}} = \frac{12}{13}$

2. Find $\sin A$ using the identity $\sin^2 A + \cos^2 A = 1$:

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

$\sin^2 A + \frac{144}{169} = 1$

$\sin^2 A = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$

Since A is an acute angle, $\sin A$ is positive.

$\sin A = \sqrt{\frac{25}{169}} = \frac{5}{13}$

3. Find $\tan A$ using the identity $\tan^2 A + 1 = \sec^2 A$ or $\tan A = \frac{\sin A}{\cos A}$. Using the identity $\tan^2 A = \sec^2 A - 1$:

$\tan^2 A = \frac{169}{144} - 1 = \frac{169 - 144}{144} = \frac{25}{144}$

Since A is an acute angle, $\tan A$ is positive.

$\tan A = \sqrt{\frac{25}{144}} = \frac{5}{12}$

4. Find $\cot A$ using the reciprocal identity $\cot A = \frac{1}{\tan A}$:

$\cot A = \frac{1}{\frac{5}{12}} = \frac{12}{5}$

5. Find $\text{cosec } A$ using the reciprocal identity $\text{cosec } A = \frac{1}{\sin A}$:

$\text{cosec } A = \frac{1}{\frac{5}{13}} = \frac{13}{5}$

---

Result:

The other trigonometric ratios for angle A are:

$\sin A = \frac{5}{13}$

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

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

$\cot A = \frac{12}{5}$

$\text{cosec } A = \frac{13}{5}$

Given:

$\tan \theta = \frac{3}{4}$

Expression to evaluate: $\frac{4 \sin \theta - \cos \theta}{4 \sin \theta + \cos \theta}$

---

Solution:

We are given the value of $\tan \theta$. We can evaluate the expression by rewriting it in terms of $\tan \theta$.

Divide the numerator and the denominator of the expression by $\cos \theta$. (Assuming $\cos \theta \neq 0$)

Separate the terms in the numerator and denominator:

$\text{Expression} = \frac{\frac{4 \sin \theta}{\cos \theta} - \frac{\cos \theta}{\cos \theta}}{\frac{4 \sin \theta}{\cos \theta} + \frac{\cos \theta}{\cos \theta}}$

Using the definition $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we substitute:

Now, substitute the given value $\tan \theta = \frac{3}{4}$ into equation (ii):

Simplify the expression:

$\text{Expression} = \frac{\cancel{4} \times \frac{3}{\cancel{4}} - 1}{\cancel{4} \times \frac{3}{\cancel{4}} + 1}$

$\text{Expression} = \frac{3 - 1}{3 + 1}$

$\text{Expression} = \frac{2}{4}$

$\text{Expression} = \frac{1}{2}$

---

Alternative Solution (Using a right triangle):

Given $\tan \theta = \frac{3}{4}$. In a right triangle, $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$.

Let the opposite side be $3k$ and the adjacent side be $4k$ for some positive constant $k$.

By the Pythagorean theorem, the hypotenuse is $\sqrt{(3k)^2 + (4k)^2} = \sqrt{9k^2 + 16k^2} = \sqrt{25k^2} = 5k$.

So, $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3k}{5k} = \frac{3}{5}$.

And $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4k}{5k} = \frac{4}{5}$.

Substitute these values into the given expression:

$\text{Expression} = \frac{4 \sin \theta - \cos \theta}{4 \sin \theta + \cos \theta} = \frac{4 \left(\frac{3}{5}\right) - \frac{4}{5}}{4 \left(\frac{3}{5}\right) + \frac{4}{5}}$

$\text{Expression} = \frac{\frac{12}{5} - \frac{4}{5}}{\frac{12}{5} + \frac{4}{5}} = \frac{\frac{12-4}{5}}{\frac{12+4}{5}} = \frac{\frac{8}{5}}{\frac{16}{5}}$

$\text{Expression} = \frac{8}{5} \times \frac{5}{16} = \frac{\cancel{8}^{1}}{\cancel{5}^{1}} \times \frac{\cancel{5}^{1}}{\cancel{16}^{2}} = \frac{1}{2}$

---

Result:

The value of the expression $\frac{4 \sin \theta - \cos \theta}{4 \sin \theta + \cos \theta}$ is $\frac{1}{2}$.

Given:

In $\triangle ABC$, right-angled at B, $\angle B = 90^\circ$.

$\tan A = 1$.

---

To Find:

The value of the expression $\sin A \cos C + \cos A \sin C$.

---

Solution:

We are given that $\tan A = 1$. For acute angle A, the value of A for which $\tan A = 1$ is $45^\circ$.

In $\triangle ABC$, which is right-angled at B, the sum of angles is $180^\circ$.

$\angle A + \angle B + \angle C = 180^\circ$

Substitute the values of A and B:

$135^\circ + C = 180^\circ$

$C = 180^\circ - 135^\circ$

Now we need to find the value of the expression $\sin A \cos C + \cos A \sin C$.

Substitute the values of A and C from (1) and (3):

The standard values are $\sin 45^\circ = \frac{1}{\sqrt{2}}$ and $\cos 45^\circ = \frac{1}{\sqrt{2}}$.

Substitute these values into the expression:

Expression = $\frac{1}{\sqrt{2} \times \sqrt{2}} + \frac{1}{\sqrt{2} \times \sqrt{2}}$

Expression = $\frac{1}{2} + \frac{1}{2}$

Expression = $\frac{1+1}{2} = \frac{2}{2} = 1$

---

Alternative Solution (Using trigonometric identity):

The expression $\sin A \cos C + \cos A \sin C$ is the expansion of the trigonometric identity for $\sin(A+C)$.

From the primary solution, we found that $A = 45^\circ$ and $C = 45^\circ$.

Therefore, $A + C = 45^\circ + 45^\circ = 90^\circ$.

Substitute $A+C = 90^\circ$ into identity (6):

The standard value of $\sin 90^\circ$ is 1.

Expression = 1

---

Result:

The value of the expression $\sin A \cos C + \cos A \sin C$ is 1.

Given:

An identity to prove: $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$.

(Assuming A is an acute angle, for which $\sec A$ and $\tan A$ are defined, and $\sin A \neq \pm 1$).

---

To Prove:

$\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$

---

Proof:

We start with the Left Hand Side (LHS) of the identity:

$\text{LHS} = \sqrt{\frac{1 + \sin A}{1 - \sin A}}$

To eliminate the square root from the denominator, we multiply the numerator and the denominator inside the square root by the conjugate of the denominator, which is $(1 + \sin A)$.

Simplify the expression inside the square root:

Using the algebraic identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:

We know the fundamental trigonometric identity $\sin^2 A + \cos^2 A = 1$. Rearranging this, we get $\cos^2 A = 1 - \sin^2 A$.

Substitute the results from (iii) and (iv) back into equation (ii):

Now, take the square root of the numerator and the denominator separately:

Since A is an acute angle ($0^\circ < A < 90^\circ$), $\sin A$ is positive and $\cos A$ is positive. Therefore, $1 + \sin A$ is positive, and $\cos A$ is positive.

So, $|1 + \sin A| = 1 + \sin A$ and $|\cos A| = \cos A$ for acute angle A.

Substitute these simplified absolute values back into equation (vi):

Separate the fraction into two terms:

Using the definitions of $\sec A$ and $\tan A$ ($\sec A = \frac{1}{\cos A}$ and $\tan A = \frac{\sin A}{\cos A}$):

This is equal to the Right Hand Side (RHS) of the identity.

$\text{LHS} = \text{RHS}$

Thus, the identity $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$ is proven for acute angles A.

---

Result:

The identity $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$ is proved.

Given:

An identity to prove: $(\sin \theta + \text{cosec } \theta)^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta$.

(Assuming $\theta$ is an angle for which all trigonometric ratios are defined).

---

To Prove:

$(\sin \theta + \text{cosec } \theta)^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta$

---

Proof:

We start with the Left Hand Side (LHS) of the identity:

$\text{LHS} = (\sin \theta + \text{cosec } \theta)^2 + (\cos \theta + \sec \theta)^2$

Expand the squared terms using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$:

Substitute equations (i) and (ii) back into the LHS expression:

Group the terms:

$\text{LHS} = (\sin^2 \theta + \cos^2 \theta) + (2 \sin \theta \text{ cosec } \theta) + (2 \cos \theta \sec \theta) + \text{cosec}^2 \theta + \sec^2 \theta$

Use the fundamental identities:

Substitute these values into the LHS expression:

Simplify the constant terms:

$\text{LHS} = 1 + 2 + 2 + \text{cosec}^2 \theta + \sec^2 \theta$

$\text{LHS} = 5 + \text{cosec}^2 \theta + \sec^2 \theta$

Now, we need to express $\text{cosec}^2 \theta$ and $\sec^2 \theta$ in terms of $\tan^2 \theta$ and $\cot^2 \theta$ using the identities:

Substitute equations (viii) and (ix) into the LHS expression:

Combine the constant terms:

$\text{LHS} = 5 + 1 + 1 + \tan^2 \theta + \cot^2 \theta$

$\text{LHS} = 7 + \tan^2 \theta + \cot^2 \theta$

This is equal to the Right Hand Side (RHS) of the identity.

$\text{LHS} = \text{RHS}$

Thus, the identity is proven.

---

Result:

The identity $(\sin \theta + \text{cosec } \theta)^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta$ is proved.

Given Expression:

$\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$

---

To Evaluate:

The value of the given expression.

---

Solution:

We need to substitute the standard values of the trigonometric ratios for $30^\circ$, $45^\circ$, and $60^\circ$.

The standard values are:

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

$\tan 45^\circ = 1$

$\text{cosec } 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$

$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$

$\cos 60^\circ = \frac{1}{2}$

$\cot 45^\circ = 1$

Substitute these values into the given expression:

Simplify the numerator and the denominator separately.

Numerator = $\frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{1+2}{2} - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}}$

To combine the terms in the numerator, find a common denominator $2\sqrt{3}$:

Numerator = $\frac{3 \times \sqrt{3}}{2 \times \sqrt{3}} - \frac{2 \times 2}{2 \times \sqrt{3}} = \frac{3\sqrt{3}}{2\sqrt{3}} - \frac{4}{2\sqrt{3}} = \frac{3\sqrt{3} - 4}{2\sqrt{3}}$

Denominator = $\frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{1+2}{2} = \frac{2}{\sqrt{3}} + \frac{3}{2}$

To combine the terms in the denominator, find a common denominator $2\sqrt{3}$:

Denominator = $\frac{2 \times 2}{2 \times \sqrt{3}} + \frac{3 \times \sqrt{3}}{2 \times \sqrt{3}} = \frac{4}{2\sqrt{3}} + \frac{3\sqrt{3}}{2\sqrt{3}} = \frac{4 + 3\sqrt{3}}{2\sqrt{3}}$

Now, substitute the simplified numerator and denominator back into the expression:

Divide the fractions by multiplying the numerator by the reciprocal of the denominator:

Expression = $\frac{3\sqrt{3} - 4}{2\sqrt{3}} \times \frac{2\sqrt{3}}{3\sqrt{3} + 4}$

Cancel out the common term $2\sqrt{3}$ (assuming $2\sqrt{3} \neq 0$):

Expression = $\frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $(3\sqrt{3} - 4)$:

Expression = $\frac{(3\sqrt{3} - 4)}{(3\sqrt{3} + 4)} \times \frac{(3\sqrt{3} - 4)}{(3\sqrt{3} - 4)}$

Using $(a-b)^2 = a^2 - 2ab + b^2$ for the numerator and $(a+b)(a-b) = a^2 - b^2$ for the denominator:

Numerator = $(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + (4)^2 = (9 \times 3) - 24\sqrt{3} + 16 = 27 - 24\sqrt{3} + 16 = 43 - 24\sqrt{3}$

Denominator = $(3\sqrt{3})^2 - (4)^2 = (9 \times 3) - 16 = 27 - 16 = 11$

Substitute the simplified numerator and denominator back into the expression:

Expression = $\frac{43 - 24\sqrt{3}}{11}$

---

Result:

The value of the expression $\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$ is $\frac{43 - 24\sqrt{3}}{11}$.

Given Formulas:

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

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

---

To Find:

The values of $\sin 15^\circ$ and $\cos 15^\circ$.

---

Solution:

We use the hint and write $15^\circ$ as the difference of two standard angles: $15^\circ = 45^\circ - 30^\circ$.

So, we can use the given formulas with $A = 45^\circ$ and $B = 30^\circ$.

The standard trigonometric values for $45^\circ$ and $30^\circ$ are:

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

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

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

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

Now, let's find $\sin 15^\circ$ using the formula $\sin (A - B) = \sin A \cos B - \cos A \sin B$ with $A = 45^\circ$ and $B = 30^\circ$:

Substitute the standard values into equation (i):

$\sin 15^\circ = \frac{1 \times \sqrt{3}}{\sqrt{2} \times 2} - \frac{1 \times 1}{\sqrt{2} \times 2}$

$\sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}$

$\sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$\sin 15^\circ = \frac{(\sqrt{3} - 1) \times \sqrt{2}}{2\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{3}\sqrt{2} - 1\sqrt{2}}{2(2)} = \frac{\sqrt{6} - \sqrt{2}}{4}$

---

Now, let's find $\cos 15^\circ$ using the formula $\cos (A - B) = \cos A \cos B + \sin A \sin B$ with $A = 45^\circ$ and $B = 30^\circ$:

Substitute the standard values into equation (iii):

$\cos 15^\circ = \frac{1 \times \sqrt{3}}{\sqrt{2} \times 2} + \frac{1 \times 1}{\sqrt{2} \times 2}$

$\cos 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$

$\cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$\cos 15^\circ = \frac{(\sqrt{3} + 1) \times \sqrt{2}}{2\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{3}\sqrt{2} + 1\sqrt{2}}{2(2)} = \frac{\sqrt{6} + \sqrt{2}}{4}$

---

Result:

The value of $\sin 15^\circ$ is $\frac{\sqrt{6} - \sqrt{2}}{4}$.

The value of $\cos 15^\circ$ is $\frac{\sqrt{6} + \sqrt{2}}{4}$.

Definitions of Trigonometric Ratios for an Acute Angle in a Right-Angled Triangle:

Consider a right-angled triangle, say $\triangle PQR$, right-angled at Q. Let $\angle P$ be an acute angle.

The sides relative to the acute angle $\angle P$ are:

Hypotenuse: The side opposite the right angle (PR).

Opposite side: The side opposite the acute angle $\angle P$ (QR).

Adjacent side: The side adjacent to the acute angle $\angle P$ (PQ).

The six trigonometric ratios are defined as follows:

1. Sine of angle P ($\sin P$) is the ratio of the length of the opposite side to the length of the hypotenuse.

$\sin P = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{QR}{PR}$

2. Cosine of angle P ($\cos P$) is the ratio of the length of the adjacent side to the length of the hypotenuse.

$\cos P = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{PQ}{PR}$

3. Tangent of angle P ($\tan P$) is the ratio of the length of the opposite side to the length of the adjacent side.

$\tan P = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{QR}{PQ}$

4. Cosecant of angle P ($\text{cosec } P$) is the reciprocal of $\sin P$.

$\text{cosec } P = \frac{1}{\sin P} = \frac{\text{Hypotenuse}}{\text{Opposite side}} = \frac{PR}{QR}$

5. Secant of angle P ($\sec P$) is the reciprocal of $\cos P$.

$\sec P = \frac{1}{\cos P} = \frac{\text{Hypotenuse}}{\text{Adjacent side}} = \frac{PR}{PQ}$

6. Cotangent of angle P ($\cot P$) is the reciprocal of $\tan P$.

$\cot P = \frac{1}{\tan P} = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{PQ}{QR}$

---

Relationship between Angles and Side Lengths:

The trigonometric ratios establish a relationship between the angles and the ratio of the lengths of sides in a right-angled triangle. For a fixed acute angle, the ratios of the sides remain constant, regardless of the size of the triangle. This means that if we know the measure of an acute angle and the length of one side, we can determine the lengths of the other two sides using these ratios. Conversely, if we know the ratio of the lengths of two sides, we can determine the measure of the acute angle (using inverse trigonometric functions).

---

Problem Calculation:

Given:

In $\triangle ABC$, right-angled at B,

AB = 24 cm

BC = 7 cm

---

To Find:

$\sin A, \cos A$, $\sin C, \cos C$.

---

Solution:

First, we need to find the length of the hypotenuse AC using the Pythagorean theorem in $\triangle ABC$:

$AC^2 = AB^2 + BC^2$

Substitute the given values of AB and BC:

$AC^2 = 576\ \text{cm}^2 + 49\ \text{cm}^2$

$AC^2 = 625\ \text{cm}^2$

Taking the square root of both sides (since AC is a length, it must be positive):

$AC = \sqrt{625\ \text{cm}^2}$

$AC = 25\ \text{cm}$

The length of the hypotenuse AC is 25 cm.

---

Now, we determine the trigonometric ratios for angle A.

For angle A:

Opposite side = BC = 7 cm

Adjacent side = AB = 24 cm

Hypotenuse = AC = 25 cm

$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC}$

$\sin A = \frac{7}{25}$

$\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC}$

$\cos A = \frac{24}{25}$

---

Next, we determine the trigonometric ratios for angle C.

For angle C:

Opposite side = AB = 24 cm

Adjacent side = BC = 7 cm

Hypotenuse = AC = 25 cm

$\sin C = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{AB}{AC}$

$\sin C = \frac{24}{25}$

$\cos C = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AC}$

$\cos C = \frac{7}{25}$

---

Result:

In $\triangle ABC$:

$\sin A = \frac{7}{25}$

$\cos A = \frac{24}{25}$

$\sin C = \frac{24}{25}$

$\cos C = \frac{7}{25}$

We will derive the values of trigonometric ratios for $45^\circ$, $30^\circ$, and $60^\circ$ using geometric constructions.

---

Trigonometric Ratios for $45^\circ$:

Consider an isosceles right-angled triangle ABC, where $\angle B = 90^\circ$ and $\angle A = \angle C = 45^\circ$.

Let the equal sides be AB = BC = $a$ units.

By the Pythagorean theorem, the hypotenuse AC can be found:

$AC^2 = AB^2 + BC^2$

$AC^2 = 2a^2$

$AC = \sqrt{2a^2} = a\sqrt{2}$ units (since length is positive).

Now, let's find the trigonometric ratios for $\angle A = 45^\circ$.

Relative to angle A:

Opposite side = BC = $a$

Adjacent side = AB = $a$

Hypotenuse = AC = $a\sqrt{2}$

$\sin 45^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\cos 45^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\tan 45^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} = \frac{a}{a} = 1$

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

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

$\cot 45^\circ = \frac{1}{\tan 45^\circ} = \frac{1}{1} = 1$

---

Trigonometric Ratios for $30^\circ$ and $60^\circ$:

Consider an equilateral triangle ABC with side length $2a$ units. All angles in an equilateral triangle are $60^\circ$, so $\angle A = \angle B = \angle C = 60^\circ$.

Draw an altitude AD from vertex A to the side BC. In an equilateral triangle, the altitude also acts as a median and an angle bisector. So, D is the midpoint of BC, and AD is perpendicular to BC.

In $\triangle ADB$, $\angle ADB = 90^\circ$.

BD = DC = $\frac{1}{2} BC = \frac{1}{2}(2a) = a$ units.

$\angle BAD = \frac{1}{2} \angle BAC = \frac{1}{2}(60^\circ) = 30^\circ$.

$\angle ABD = \angle B = 60^\circ$.

Now, consider the right-angled triangle ADB with angles $30^\circ, 60^\circ, 90^\circ$ and sides AB = $2a$, BD = $a$. By the Pythagorean theorem, we find the length of the altitude AD:

$AD^2 + BD^2 = AB^2$

$AD^2 + a^2 = 4a^2$

$AD^2 = 4a^2 - a^2 = 3a^2$

$AD = \sqrt{3a^2} = a\sqrt{3}$ units (since length is positive).

Now we have the side lengths of the right triangle ADB relative to the angles $30^\circ$ and $60^\circ$:

Hypotenuse = AB = $2a$

Side opposite $30^\circ$ (BD) = $a$

Side adjacent to $30^\circ$ (AD) = $a\sqrt{3}$

Side opposite $60^\circ$ (AD) = $a\sqrt{3}$

Side adjacent to $60^\circ$ (BD) = $a$

Let's find the trigonometric ratios for $\angle BAD = 30^\circ$:

$\sin 30^\circ = \frac{\text{Opposite (BD)}}{\text{Hypotenuse (AB)}} = \frac{a}{2a} = \frac{1}{2}$

$\cos 30^\circ = \frac{\text{Adjacent (AD)}}{\text{Hypotenuse (AB)}} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2}$

$\tan 30^\circ = \frac{\text{Opposite (BD)}}{\text{Adjacent (AD)}} = \frac{a}{a\sqrt{3}} = \frac{1}{\sqrt{3}}$

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

$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$

$\cot 30^\circ = \frac{1}{\tan 30^\circ} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}$

Now, let's find the trigonometric ratios for $\angle ABD = 60^\circ$:

$\sin 60^\circ = \frac{\text{Opposite (AD)}}{\text{Hypotenuse (AB)}} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2}$

$\cos 60^\circ = \frac{\text{Adjacent (BD)}}{\text{Hypotenuse (AB)}} = \frac{a}{2a} = \frac{1}{2}$

$\tan 60^\circ = \frac{\text{Opposite (AD)}}{\text{Adjacent (BD)}} = \frac{a\sqrt{3}}{a} = \sqrt{3}$

$\text{cosec } 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$

$\sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{\frac{1}{2}} = 2$

$\cot 60^\circ = \frac{1}{\tan 60^\circ} = \frac{1}{\sqrt{3}}$

---

Summary of Values:

Angle $\theta$	$\sin \theta$	$\cos \theta$	$\tan \theta$	$\text{cosec } \theta$	$\sec \theta$	$\cot \theta$
$30^\circ$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$	2	$\frac{2}{\sqrt{3}}$	$\sqrt{3}$
$45^\circ$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	1	$\sqrt{2}$	$\sqrt{2}$	1
$60^\circ$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$	$\frac{2}{\sqrt{3}}$	2	$\frac{1}{\sqrt{3}}$

---

Result:

The values of the trigonometric ratios for $30^\circ$, $45^\circ$, and $60^\circ$ have been derived using geometric methods and summarized in the table above.

Proof of Fundamental Identity: $\sin^2 \theta + \cos^2 \theta = 1$

Given:

An identity to prove: $\sin^2 \theta + \cos^2 \theta = 1$.

We will prove this using a right-angled triangle for an acute angle $\theta$.

---

To Prove:

$\sin^2 \theta + \cos^2 \theta = 1$

---

Proof:

Consider a right-angled triangle ABC, right-angled at B. Let $\angle A = \theta$, where $\theta$ is an acute angle.

In $\triangle ABC$, let the side lengths opposite to vertices A, B, C be $a, b, c$ respectively. So, BC = $a$, AC = $b$, AB = $c$.

By the definition of trigonometric ratios for angle $\theta$ (angle A):

According to the Pythagorean theorem in the right-angled triangle ABC:

Now, consider the Left Hand Side (LHS) of the identity: $\sin^2 \theta + \cos^2 \theta$.

Substitute the expressions for $\sin \theta$ and $\cos \theta$ from equations (1) and (2) into the LHS:

Simplify the expression:

$\text{LHS} = \frac{a^2}{b^2} + \frac{c^2}{b^2}$

$\text{LHS} = \frac{a^2 + c^2}{b^2}$

From the Pythagorean theorem (equation 4), we know that $c^2 + a^2 = b^2$. Substitute this into the numerator of the expression for LHS:

Assuming $b \neq 0$ (which is true for a triangle), we can cancel out $b^2$ from the numerator and denominator:

$\text{LHS} = 1$

The Right Hand Side (RHS) of the identity is also 1.

Since LHS = RHS, the identity $\sin^2 \theta + \cos^2 \theta = 1$ is proven for an acute angle $\theta$. This identity is fundamental and holds for all real values of $\theta$ where $\sin \theta$ and $\cos \theta$ are defined.

---

Proof of Identity: $\sec^2 \theta - \tan^2 \theta = 1$

Given:

The fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$.

An identity to prove: $\sec^2 \theta - \tan^2 \theta = 1$.

(Assuming $\cos \theta \neq 0$).

---

To Prove:

$\sec^2 \theta - \tan^2 \theta = 1$

---

Proof:

Start with the fundamental trigonometric identity:

Assuming $\cos \theta \neq 0$, divide both sides of equation (F1) by $\cos^2 \theta$:

Use the definitions of $\tan \theta$ and $\sec \theta$:

$\frac{\sin^2 \theta}{\cos^2 \theta} = \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \tan^2 \theta$

$\frac{\cos^2 \theta}{\cos^2 \theta} = 1$

$\frac{1}{\cos^2 \theta} = \left(\frac{1}{\cos \theta}\right)^2 = \sec^2 \theta$

Substitute these into equation (P2.1):

Rearrange equation (P2.2) to match the required identity by subtracting $\tan^2 \theta$ from both sides:

$1 = \sec^2 \theta - \tan^2 \theta$

Or, writing it in the standard form:

$\sec^2 \theta - \tan^2 \theta = 1$

Thus, the identity $\sec^2 \theta - \tan^2 \theta = 1$ is proven for angles where $\cos \theta \neq 0$.

---

Proof of Identity: $\text{cosec}^2 \theta - \cot^2 \theta = 1$

Given:

The fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$.

An identity to prove: $\text{cosec}^2 \theta - \cot^2 \theta = 1$.

(Assuming $\sin \theta \neq 0$).

---

To Prove:

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

---

Proof:

Start with the fundamental trigonometric identity:

Assuming $\sin \theta \neq 0$, divide both sides of equation (F1) by $\sin^2 \theta$:

Use the definitions of $\cot \theta$ and $\text{cosec } \theta$:

$\frac{\sin^2 \theta}{\sin^2 \theta} = 1$

$\frac{\cos^2 \theta}{\sin^2 \theta} = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \cot^2 \theta$

$\frac{1}{\sin^2 \theta} = \left(\frac{1}{\sin \theta}\right)^2 = \text{cosec}^2 \theta$

Substitute these into equation (P3.1):

Rearrange equation (P3.2) to match the required identity by subtracting $\cot^2 \theta$ from both sides:

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

Or, writing it in the standard form:

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

Thus, the identity $\text{cosec}^2 \theta - \cot^2 \theta = 1$ is proven for angles where $\sin \theta \neq 0$.

---

Result:

The three fundamental trigonometric identities $\sin^2 \theta + \cos^2 \theta = 1$, $\sec^2 \theta - \tan^2 \theta = 1$, and $\text{cosec}^2 \theta - \cot^2 \theta = 1$ have been proved.

Proof of Identity (a):

$\frac{(\sin \theta - 2 \sin^3 \theta)}{(2 \cos^3 \theta - \cos \theta)} = \tan \theta$

Given:

An identity to prove: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$.

(Assuming $\cos \theta \neq 0$ and the denominator is non-zero).

---

To Prove:

$\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$

---

Proof:

Start with the Left Hand Side (LHS):

$\text{LHS} = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$

Factor out common terms from the numerator and the denominator:

We know the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$.

From this, $\sin^2 \theta = 1 - \cos^2 \theta$. Substitute this into the term $(1 - 2 \sin^2 \theta)$ in the numerator:

$1 - 2 \sin^2 \theta = 1 - 2 + 2 \cos^2 \theta$

Now substitute the result from equation (3) back into the numerator of equation (1):

Assuming $(2 \cos^2 \theta - 1) \neq 0$, we can cancel out the term $(2 \cos^2 \theta - 1)$ from the numerator and the denominator:

Using the definition $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

This is equal to the Right Hand Side (RHS) of the identity.

$\text{LHS} = \text{RHS}$

Thus, the identity $\frac{(\sin \theta - 2 \sin^3 \theta)}{(2 \cos^3 \theta - \cos \theta)} = \tan \theta$ is proven.

---

Proof of Identity (b):

$(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$

Given:

An identity to prove: $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$.

(Assuming A is an angle for which all trigonometric ratios are defined).

---

To Prove:

$(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$

---

Proof:

We start with the Left Hand Side (LHS) of the identity:

$\text{LHS} = (\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2$

Expand the squared terms using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$:

Substitute equations (i) and (ii) back into the LHS expression:

Group the terms:

$\text{LHS} = (\sin^2 A + \cos^2 A) + (2 \sin A \text{ cosec } A) + (2 \cos A \sec A) + \text{cosec}^2 A + \sec^2 A$

Use the fundamental identities:

Substitute these values into the LHS expression:

Simplify the constant terms:

$\text{LHS} = 1 + 2 + 2 + \text{cosec}^2 A + \sec^2 A$

$\text{LHS} = 5 + \text{cosec}^2 A + \sec^2 A$

Now, we express $\text{cosec}^2 A$ and $\sec^2 A$ in terms of $\tan^2 A$ and $\cot^2 A$ using the identities derived from the fundamental one:

Substitute equations (viii) and (ix) into the LHS expression:

Combine the constant terms:

$\text{LHS} = 5 + 1 + 1 + \tan^2 A + \cot^2 A$

$\text{LHS} = 7 + \tan^2 A + \cot^2 A$

This is equal to the Right Hand Side (RHS) of the identity.

$\text{LHS} = \text{RHS}$

Thus, the identity $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$ is proven.

---

Result:

Both trigonometric identities have been proved.

Given:

$\tan (A + B) = \sqrt{3}$

$\tan (A - B) = \frac{1}{\sqrt{3}}$

Conditions: $0^\circ < A + B \leq 90^\circ$ and $A > B$.

---

To Find:

The values of A and B.

---

Solution:

We are given $\tan (A + B) = \sqrt{3}$. We know that $\tan 60^\circ = \sqrt{3}$.

Given the condition $0^\circ < A + B \leq 90^\circ$, the angle $A+B$ must be $60^\circ$.

We are also given $\tan (A - B) = \frac{1}{\sqrt{3}}$. We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

Given that $A > B$, $A - B$ must be a positive angle. The positive value of $\tan(A-B)$ suggests that $A-B$ is in the first quadrant. Therefore, the angle $A-B$ must be $30^\circ$.

Now we have a system of two linear equations for A and B:

Equation (1): $A + B = 60^\circ$

Equation (2): $A - B = 30^\circ$

Add Equation (1) and Equation (2):

$A + B + A - B = 90^\circ$

$2A = 90^\circ$

$A = \frac{90^\circ}{2}$

$A = 45^\circ$

Substitute the value of A into Equation (1):

$B = 60^\circ - 45^\circ$

$B = 15^\circ$

---

Verification:

Check the conditions:

$0^\circ < A + B \leq 90^\circ$:

$A + B = 45^\circ + 15^\circ = 60^\circ$. Since $0^\circ < 60^\circ \leq 90^\circ$, this condition is satisfied.

$A > B$:

$A = 45^\circ$ and $B = 15^\circ$. Since $45^\circ > 15^\circ$, this condition is satisfied.

Check the original equations:

$\tan (A + B) = \tan (45^\circ + 15^\circ) = \tan 60^\circ = \sqrt{3}$ (Matches given).

$\tan (A - B) = \tan (45^\circ - 15^\circ) = \tan 30^\circ = \frac{1}{\sqrt{3}}$ (Matches given).

All conditions and equations are satisfied by $A = 45^\circ$ and $B = 15^\circ$.

---

Result:

The values of A and B are $A = 45^\circ$ and $B = 15^\circ$.

This problem statement typically assumes that the angles of elevation from two points are given to be complementary, and asks to prove the height of the tower based on this condition and the distances of the points from the base. We will interpret the question in this standard way: prove the height is $\sqrt{ab}$, *given* that the angles of elevation from two points at distances $a$ and $b$ (on the same straight line with the base and on the same side) are complementary.

---

Given:

Let TB be a tower of height $h$, where B is the base and T is the top.

Let P1 and P2 be two points on the ground in the same straight line with the base B, and on the same side of the tower.

Let BP1 = $a$ and BP2 = $b$. Assume P1 is closer to the base, so $a < b$.

Let the angle of elevation of the top of the tower T from P1 be $\alpha$, i.e., $\angle TP1B = \alpha$.

Let the angle of elevation of the top of the tower T from P2 be $\beta$, i.e., $\angle TP2B = \beta$.

The angles of elevation are complementary, which means their sum is $90^\circ$.

---

To Prove:

The height of the tower, $h$, is $\sqrt{ab}$.

---

Proof:

Consider the right-angled triangle $\triangle TBP1$. The angle of elevation from P1 is $\alpha$.

Using the definition of the tangent ratio:

Consider the right-angled triangle $\triangle TBP2$. The angle of elevation from P2 is $\beta$.

Using the definition of the tangent ratio:

From the given information (Equation 1), we have $\alpha + \beta = 90^\circ$. This implies $\beta = 90^\circ - \alpha$.

Substitute this into Equation (3):

Using the trigonometric identity for complementary angles, $\tan (90^\circ - \theta) = \cot \theta$:

Substitute Equation (5) into Equation (4):

We also know the reciprocal identity $\cot \theta = \frac{1}{\tan \theta}$. So, $\cot \alpha = \frac{1}{\tan \alpha}$.

Equating the expressions for $\cot \alpha$ from Equation (6) and Equation (7):

Now substitute the expression for $\tan \alpha$ from Equation (2) into Equation (8):

Simplify the right side of Equation (9): $\frac{1}{\frac{h}{a}} = \frac{a}{h}$.

So, we have:

Cross-multiply in Equation (10):

$h \times h = a \times b$

$h^2 = ab$

Taking the square root of both sides (since $h$ is a height, it must be positive):

$h = \sqrt{ab}$

This proves that the height of the tower is $\sqrt{ab}$.

Note: The phrasing "Prove that the angle of elevation ... are complementary" based on the distances $a$ and $b$ requires additional assumptions (e.g., specific values for $a, b, h$) or a different problem setup. Assuming the standard problem structure, we proved that *if* the angles are complementary, the height is $\sqrt{ab}$.

---

Result:

Assuming the angles of elevation from two points at distances $a$ and $b$ from the base of the tower (on the same straight line and same side) are complementary, the height of the tower is proved to be $\sqrt{ab}$.

Given:

An identity to prove: $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$.

Identity to be used: $\sec^2 \theta = 1 + \tan^2 \theta$, which can be rearranged as $\sec^2 \theta - \tan^2 \theta = 1$.

(Assuming $\cos \theta \neq 0$ and the denominator is non-zero).

---

To Prove:

$\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$

---

Proof:

We start with the Left Hand Side (LHS) of the identity:

$\text{LHS} = \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}$

To introduce $\tan \theta$ and $\sec \theta$, divide the numerator and the denominator by $\cos \theta$ (assuming $\cos \theta \neq 0$):

Use the definitions $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$:

Rearrange the terms in the numerator for clarity:

We need to use the identity $\sec^2 \theta - \tan^2 \theta = 1$. Let's replace the constant term '1' in the denominator with this identity.

Factor the difference of squares term in the denominator: $\sec^2 \theta - \tan^2 \theta = (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)$.

Factor out the common term $(\sec \theta - \tan \theta)$ from the terms in the denominator:

The term in the square brackets in the denominator is $(\sec \theta + \tan \theta) - 1 = \sec \theta + \tan \theta - 1$.

The numerator is $\tan \theta + \sec \theta - 1$, which is the same expression.

Assuming $\sec \theta + \tan \theta - 1 \neq 0$, we can cancel the common term from the numerator and the denominator:

This is equal to the Right Hand Side (RHS) of the identity.

$\text{LHS} = \text{RHS}$

Thus, the identity $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$ is proven using the identity $\sec^2 \theta = 1 + \tan^2 \theta$.

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Result:

The identity is proved.

Concept of Trigonometric Ratios of Complementary Angles:

Two angles are said to be complementary if their sum is $90^\circ$.

In a right-angled triangle, the right angle is $90^\circ$. The sum of all angles in a triangle is $180^\circ$. Therefore, in a right-angled triangle, the sum of the other two acute angles must be $180^\circ - 90^\circ = 90^\circ$. This means the two acute angles in a right-angled triangle are always complementary to each other.

For example, if one acute angle is A, the other acute angle is $90^\circ - A$.

The trigonometric ratios of one acute angle in a right triangle are related to the trigonometric co-ratios of its complementary angle. Specifically, sine of an angle is equal to the cosine of its complementary angle, tangent of an angle is equal to the cotangent of its complementary angle, and secant of an angle is equal to the cosecant of its complementary angle, and vice-versa.

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Proof of Identities:

Given:

Identities to prove: $\sin (90^\circ - A) = \cos A$ and $\tan (90^\circ - A) = \cot A$.

(Assuming A is an acute angle, $0^\circ < A < 90^\circ$).

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To Prove:

$\sin (90^\circ - A) = \cos A$

$\tan (90^\circ - A) = \cot A$

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Proof:

Consider a right-angled triangle ABC, right-angled at B. Let $\angle BAC = A$.

Since $\angle ABC = 90^\circ$ and the sum of angles in a triangle is $180^\circ$, we have:

$\angle BAC + \angle ABC + \angle BCA = 180^\circ$

$A + 90^\circ + \angle BCA = 180^\circ$

$\angle BCA = 180^\circ - 90^\circ - A$

Let's denote $\angle C = 90^\circ - A$. So, A and C are complementary angles.

By the definition of trigonometric ratios in $\triangle ABC$:

For angle A:

For angle C ($90^\circ - A$):

Comparing equations (6) and (3):

$\sin (90^\circ - A) = \frac{AB}{AC}$ and $\cos A = \frac{AB}{AC}$.

(Note: Similarly, comparing (7) and (2) shows $\cos (90^\circ - A) = \sin A$).

Comparing equations (8) and (5):

$\tan (90^\circ - A) = \frac{AB}{BC}$ and $\cot A = \frac{AB}{BC}$.

(Note: Similarly, comparing (9) and (4) shows $\cot (90^\circ - A) = \tan A$).

Other complementary identities can be derived similarly: $\sec (90^\circ - A) = \text{cosec } A$ and $\text{cosec } (90^\circ - A) = \sec A$.

Thus, the identities are proven for acute angle A.

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Evaluation of Expression:

Given Expression:

$\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ$

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To Evaluate:

The value of the given expression.

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Solution:

We group the angles that are complementary:

$48^\circ + 42^\circ = 90^\circ$

$23^\circ + 67^\circ = 90^\circ$

Use the identity $\tan (90^\circ - \theta) = \cot \theta$.

Let $\theta = 42^\circ$. Then $90^\circ - 42^\circ = 48^\circ$. So, $\tan 48^\circ = \tan (90^\circ - 42^\circ) = \cot 42^\circ$.

Let $\theta = 67^\circ$. Then $90^\circ - 67^\circ = 23^\circ$. So, $\tan 23^\circ = \tan (90^\circ - 67^\circ) = \cot 67^\circ$.

Substitute these into the expression:

Rearrange the terms to group tangent and cotangent of the same angle:

Use the reciprocal identity $\tan \theta \cot \theta = 1$.

$\cot 42^\circ \tan 42^\circ = 1$ (assuming $\tan 42^\circ \neq 0$ and $\cot 42^\circ \neq 0$)

$\cot 67^\circ \tan 67^\circ = 1$ (assuming $\tan 67^\circ \neq 0$ and $\cot 67^\circ \neq 0$)

Substitute these values into equation (Ev2):

Expression = $1$

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Result:

The concept of trigonometric ratios of complementary angles is explained. The identities $\sin (90^\circ - A) = \cos A$ and $\tan (90^\circ - A) = \cot A$ are proved.

The value of the expression $\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ$ is 1.

Given:

An identity to prove: $\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\text{cosec } A - 1}{\text{cosec } A + 1}$.

(Assuming $\sin A \neq 0$, $\cos A \neq 0$, and the denominators are non-zero).

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To Prove:

$\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\text{cosec } A - 1}{\text{cosec } A + 1}$

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Proof:

We start with the Left Hand Side (LHS) of the identity:

$\text{LHS} = \frac{\cot A - \cos A}{\cot A + \cos A}$

Use the definition of cotangent: $\cot A = \frac{\cos A}{\sin A}$. Substitute this into the LHS:

Factor out $\cos A$ from the numerator and the denominator:

Assuming $\cos A \neq 0$, cancel out the common term $\cos A$ from the numerator and the denominator:

Use the definition of cosecant: $\text{cosec } A = \frac{1}{\sin A}$. Substitute this into equation (iii):

This is equal to the Right Hand Side (RHS) of the identity.

$\text{LHS} = \text{RHS}$

Thus, the identity $\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\text{cosec } A - 1}{\text{cosec } A + 1}$ is proven.

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Result:

The identity $\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\text{cosec } A - 1}{\text{cosec } A + 1}$ is proved.

Given:

$\sin \theta + \cos \theta = \sqrt{2} \cos \theta$

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To Prove:

$\cos \theta - \sin \theta = \sqrt{2} \sin \theta$

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Proof:

We start with the given equation:

Rearrange the equation to isolate the $\sin \theta$ term on one side:

Factor out $\cos \theta$ from the terms on the right side of equation (2):

Now, multiply both sides of equation (3) by the conjugate of $(\sqrt{2} - 1)$, which is $(\sqrt{2} + 1)$. This step is performed to rationalize the term involving $\sqrt{2}-1$ and simplify the expression.

Expand the left side of equation (4):

$\sqrt{2} \sin \theta + \sin \theta$

Expand the right side of equation (4) using the algebraic identity $(a-b)(a+b) = a^2 - b^2$:

$(\sqrt{2} - 1)(\sqrt{2} + 1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1$

So, the right side becomes $\cos \theta (1) = \cos \theta$.

Equating the expanded left and right sides of equation (4):

Rearrange equation (5) to match the expression we need to prove by subtracting $\sin \theta$ from both sides:

Rewrite equation (6) in the desired form:

$\cos \theta - \sin \theta = \sqrt{2} \sin \theta$

This matches the expression we were required to prove.

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Result:

Given $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$, it is proved that $\cos \theta - \sin \theta = \sqrt{2} \sin \theta$.

Given Expression:

$\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$

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To Evaluate:

The value of the given expression.

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Solution:

We need to substitute the exact standard values of the trigonometric ratios for $60^\circ$, $30^\circ$, and $45^\circ$.

The standard values are:

$\cos 60^\circ = \frac{1}{2}$

$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$

$\tan 45^\circ = 1$

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

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

Substitute these values into the given expression:

Now, evaluate the squared terms:

$\left(\frac{1}{2}\right)^2 = \frac{1^2}{2^2} = \frac{1}{4}$

$\left(\frac{2}{\sqrt{3}}\right)^2 = \frac{2^2}{(\sqrt{3})^2} = \frac{4}{3}$

$(1)^2 = 1$

$\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{(\sqrt{3})^2}{2^2} = \frac{3}{4}$

Substitute the squared values back into the expression (i):

Perform the multiplications in the numerator:

$5 \left(\frac{1}{4}\right) = \frac{5}{4}$

$4 \left(\frac{4}{3}\right) = \frac{16}{3}$

Substitute these back into the numerator of expression (ii):

Numerator = $\frac{5}{4} + \frac{16}{3} - 1$

To simplify the numerator, find a common denominator for the terms, which is 12:

Numerator = $\frac{5 \times 3}{4 \times 3} + \frac{16 \times 4}{3 \times 4} - \frac{1 \times 12}{1 \times 12}$

Numerator = $\frac{15}{12} + \frac{64}{12} - \frac{12}{12}$

Numerator = $\frac{15 + 64 - 12}{12} = \frac{79 - 12}{12} = \frac{67}{12}$

Now, simplify the denominator of expression (ii):

Denominator = $\frac{1}{4} + \frac{3}{4} = \frac{1 + 3}{4} = \frac{4}{4} = 1$

Substitute the simplified numerator and denominator back into expression (ii):

Expression = $\frac{67}{12}$

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Alternative Calculation for Denominator:

The denominator is $\sin^2 30^\circ + \cos^2 30^\circ$.

Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, with $\theta = 30^\circ$:

This directly gives the denominator value as 1, simplifying the calculation as shown above.

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Result:

The value of the expression $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$ is $\frac{67}{12}$.

Given:

An identity to prove: $(\sec A - \cos A)(\cot A + \tan A) = \tan A \sec A$.

(Assuming A is an angle for which all trigonometric ratios are defined, i.e., $\sin A \neq 0$, $\cos A \neq 0$).

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To Prove:

$(\sec A - \cos A)(\cot A + \tan A) = \tan A \sec A$

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Proof:

We start with the Left Hand Side (LHS) of the identity:

$\text{LHS} = (\sec A - \cos A)(\cot A + \tan A)$

Express the trigonometric ratios in terms of $\sin A$ and $\cos A$:

$\sec A = \frac{1}{\cos A}$

$\cot A = \frac{\cos A}{\sin A}$

$\tan A = \frac{\sin A}{\cos A}$

Substitute these definitions into the LHS expression:

Simplify the terms within the first parenthesis by finding a common denominator:

Using the fundamental identity $\sin^2 A + \cos^2 A = 1$, we know that $1 - \cos^2 A = \sin^2 A$. Substitute this into equation (ii):

Now simplify the terms within the second parenthesis by finding a common denominator ($\sin A \cos A$):

Using the fundamental identity $\cos^2 A + \sin^2 A = 1$, substitute into equation (iv):

Substitute the simplified forms from equations (iii) and (v) back into the LHS expression (equation i):

Multiply the terms:

Cancel out one factor of $\sin A$ from the numerator and denominator (assuming $\sin A \neq 0$):

Now, consider the Right Hand Side (RHS) of the identity:

$\text{RHS} = \tan A \sec A$

Express the terms in the RHS in terms of $\sin A$ and $\cos A$:

Multiply the terms:

Comparing equation (ix) and equation (xi), we see that LHS = RHS.

$\frac{\sin A}{\cos^2 A} = \frac{\sin A}{\cos^2 A}$

Thus, the identity $(\sec A - \cos A)(\cot A + \tan A) = \tan A \sec A$ is proven.

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Result:

The identity $(\sec A - \cos A)(\cot A + \tan A) = \tan A \sec A$ is proved.