Chapter 8 Introduction To Trigonometry

Given:

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

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To Find:

The other trigonometric ratios of the angle A.

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Solution:

Consider a right-angled triangle ABC, right-angled at B, with respect to angle A.

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

$\tan A = \frac{\text{Side opposite to angle A}}{\text{Side adjacent to angle A}} = \frac{BC}{AB}$

We are given $\tan A = \frac{4}{3}$.

So, $\frac{BC}{AB} = \frac{4}{3}$.

Let $BC = 4k$ and $AB = 3k$, where $k$ is a positive constant.

By the Pythagorean theorem in $\triangle$ ABC:

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

$AC^2 = (3k)^2 + (4k)^2$

$AC^2 = 9k^2 + 16k^2$

$AC^2 = 25k^2$

$AC = \sqrt{25k^2} = 5k$ (Since length is positive)

Now we can find the other trigonometric ratios:

$\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{4k}{5k} = \frac{4}{5}$

$\cos A = \frac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{3k}{5k} = \frac{3}{5}$

$\text{cosec } A = \frac{1}{\sin A} = \frac{\text{Hypotenuse}}{\text{Side opposite to angle A}} = \frac{AC}{BC} = \frac{5k}{4k} = \frac{5}{4}$

$\sec A = \frac{1}{\cos A} = \frac{\text{Hypotenuse}}{\text{Side adjacent to angle A}} = \frac{AC}{AB} = \frac{5k}{3k} = \frac{5}{3}$

$\cot A = \frac{1}{\tan A} = \frac{\text{Side adjacent to angle A}}{\text{Side opposite to angle A}} = \frac{AB}{BC} = \frac{3k}{4k} = \frac{3}{4}$

The other trigonometric ratios of angle A are:

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

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

$\text{cosec } A = \frac{5}{4}$

$\sec A = \frac{5}{3}$

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

Given:

$\angle$B and $\angle$Q are acute angles.

$\sin$ B = $\sin$ Q.

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To Prove:

$\angle$B = $\angle$Q.

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Solution:

Consider two right-angled triangles, $\triangle$ ABC and $\triangle$ PQR, where $\angle$C is right-angled in $\triangle$ ABC and $\angle$R is right-angled in $\triangle$ PQR. Let $\angle$B be in $\triangle$ ABC and $\angle$Q be in $\triangle$ PQR, and both $\angle$B and $\angle$Q are acute angles.

By the definition of the sine ratio:

In $\triangle$ ABC, $\sin$ B = $\frac{\text{Side opposite to }\angle\text{B}}{\text{Hypotenuse}} = \frac{AC}{AB}$

In $\triangle$ PQR, $\sin$ Q = $\frac{\text{Side opposite to }\angle\text{Q}}{\text{Hypotenuse}} = \frac{PR}{PQ}$

We are given that $\sin$ B = $\sin$ Q.

Let $\frac{AC}{AB} = \frac{PR}{PQ} = k$, where $k$ is a positive constant.

Then, $AC = k \cdot AB$ and $PR = k \cdot PQ$.

Using the Pythagorean theorem in $\triangle$ ABC:

$BC^2 = AB^2 - AC^2$

$BC = \sqrt{AB^2 - AC^2} = \sqrt{AB^2 - (k \cdot AB)^2} = \sqrt{AB^2 - k^2 AB^2} = \sqrt{AB^2(1 - k^2)} = AB \sqrt{1 - k^2}$

Using the Pythagorean theorem in $\triangle$ PQR:

$QR^2 = PQ^2 - PR^2$

$QR = \sqrt{PQ^2 - PR^2} = \sqrt{PQ^2 - (k \cdot PQ)^2} = \sqrt{PQ^2 - k^2 PQ^2} = \sqrt{PQ^2(1 - k^2)} = PQ \sqrt{1 - k^2}$

Now consider the ratio of the other pair of corresponding sides, BC and QR:

$\frac{BC}{QR} = \frac{AB \sqrt{1 - k^2}}{PQ \sqrt{1 - k^2}}$

Assuming $1 - k^2 \neq 0$ (which must be true for acute angles, as $k = \sin B = \sin Q$ and $\sin$ of an acute angle is between 0 and 1), we can cancel $\sqrt{1 - k^2}$.

$\frac{BC}{QR} = \frac{AB}{PQ}$

From equation (i), we have $\frac{AC}{PR} = \frac{AB}{PQ}$ (by rearranging the terms).

So, we have the ratios of corresponding sides equal:

Since the ratios of all three pairs of corresponding sides are equal, $\triangle$ ABC is similar to $\triangle$ PQR by the SSS similarity criterion.

When two triangles are similar, their corresponding angles are equal.

Therefore, $\angle$B = $\angle$Q.

Hence Proved.

Given:

In $\triangle$ ACB, $\angle$C = $90^\circ$, AB = 29 units, BC = 21 units, and $\angle$ABC = $\theta$.

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To Find:

(i) $\cos^2 \theta + \sin^2 \theta$

(ii) $\cos^2 \theta - \sin^2 \theta$

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Solution:

In the right-angled $\triangle$ ACB, by the Pythagorean theorem, we have:

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

$(29)^2 = AC^2 + (21)^2$

$841 = AC^2 + 441$

$AC^2 = 841 - 441$

$AC^2 = 400$

$AC = \sqrt{400} = 20$ units (Since length is positive)

Now, we determine the trigonometric ratios $\sin \theta$ and $\cos \theta$ with respect to angle $\theta$ ($\angle$ABC):

$\sin \theta = \frac{\text{Side opposite to }\theta}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{20}{29}$

$\cos \theta = \frac{\text{Side adjacent to }\theta}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{21}{29}$

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(i) Value of $\cos^2 \theta + \sin^2 \theta$:

$\cos^2 \theta + \sin^2 \theta = \left(\frac{21}{29}\right)^2 + \left(\frac{20}{29}\right)^2$

$= \frac{21^2}{29^2} + \frac{20^2}{29^2}$

$= \frac{441}{841} + \frac{400}{841}$

$= \frac{441 + 400}{841}$

$= \frac{841}{841} = 1$

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

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(ii) Value of $\cos^2 \theta - \sin^2 \theta$:

$\cos^2 \theta - \sin^2 \theta = \left(\frac{21}{29}\right)^2 - \left(\frac{20}{29}\right)^2$

$= \frac{21^2}{29^2} - \frac{20^2}{29^2}$

$= \frac{441}{841} - \frac{400}{841}$

$= \frac{441 - 400}{841}$

$= \frac{41}{841}$

So, $\cos^2 \theta - \sin^2 \theta = \frac{41}{841}$.

Given:

In $\triangle$ ABC, $\angle$B = $90^\circ$ and $\tan A = 1$.

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To Verify:

$2 \sin A \cos A = 1$.

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Solution:

In a right triangle ABC, right-angled at B, the trigonometric ratio $\tan A$ is defined as the ratio of the side opposite to angle A to the side adjacent to angle A.

$\tan A = \frac{\text{Side opposite to angle A}}{\text{Side adjacent to angle A}} = \frac{BC}{AB}$

We are given $\tan A = 1$.

So, $\frac{BC}{AB} = 1$, which implies $BC = AB$.

Let $AB = BC = k$, where $k$ is a positive real number.

By the Pythagorean theorem in $\triangle$ ABC, the hypotenuse AC is:

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

$AC^2 = k^2 + k^2$

$AC^2 = 2k^2$

$AC = \sqrt{2k^2} = k\sqrt{2}$ (Since length is positive)

Now, we find the values of $\sin A$ and $\cos A$:

$\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\cos A = \frac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}$

Now, we evaluate the expression $2 \sin A \cos A$:

$2 \sin A \cos A = 2 \times \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{1}{\sqrt{2}}\right)$

$= 2 \times \frac{1}{\sqrt{2} \times \sqrt{2}}$

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

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

Thus, we have verified that $2 \sin A \cos A = 1$ when $\tan A = 1$ in a right-angled triangle at B.

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Alternate Solution:

We are given $\tan A = 1$.

We know that $\tan 45^\circ = 1$. Since A is an acute angle in a right triangle, this implies that $A = 45^\circ$.

Now, we find $\sin A$ and $\cos A$ for $A = 45^\circ$:

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

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

Substitute these values into the expression $2 \sin A \cos A$:

$2 \sin A \cos A = 2 \times \sin 45^\circ \times \cos 45^\circ$

$= 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$

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

$= 1$

Thus, $2 \sin A \cos A = 1$ is verified.

Given:

In $\triangle$ OPQ, $\angle$P = $90^\circ$, OP = 7 cm and OQ – PQ = 1 cm.

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To Find:

The values of $\sin$ Q and $\cos$ Q.

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Solution:

In the right-angled $\triangle$ OPQ, we are given:

From equation (i), we can express OQ in terms of PQ:

Applying the Pythagorean theorem to $\triangle$ OPQ:

$OQ^2 = OP^2 + PQ^2$

Substitute the values from the given information and the relation between OQ and PQ:

$(PQ + 1)^2 = (7)^2 + PQ^2$

Expand the left side using $(a+b)^2 = a^2 + 2ab + b^2$:

$PQ^2 + 2(PQ)(1) + 1^2 = 49 + PQ^2$

$PQ^2 + 2PQ + 1 = 49 + PQ^2$

Subtract $PQ^2$ from both sides:

$2PQ + 1 = 49$

Subtract 1 from both sides:

$2PQ = 49 - 1$

$2PQ = 48$

Divide by 2:

$PQ = \frac{48}{2}$

$PQ = 24$ cm

Now substitute the value of PQ into the relation OQ = PQ + 1:

$OQ = 24 + 1$

$OQ = 25$ cm

So, the lengths of the sides are OP = 7 cm (Opposite to angle Q), PQ = 24 cm (Adjacent to angle Q), and OQ = 25 cm (Hypotenuse).

Now, we can find the values of $\sin$ Q and $\cos$ Q:

$\sin Q = \frac{\text{Side opposite to angle Q}}{\text{Hypotenuse}} = \frac{OP}{OQ}$

$\sin Q = \frac{7}{25}$

$\cos Q = \frac{\text{Side adjacent to angle Q}}{\text{Hypotenuse}} = \frac{PQ}{OQ}$

$\cos Q = \frac{24}{25}$

The values are $\sin Q = \frac{7}{25}$ and $\cos Q = \frac{24}{25}$.

Given:

In $\triangle$ ABC, $\angle$B = $90^\circ$, AB = 24 cm, BC = 7 cm.

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To Find:

(i) $\sin$ A, $\cos$ A

(ii) $\sin$ C, $\cos$ C

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Solution:

In the right-angled $\triangle$ ABC, by the Pythagorean theorem, we have:

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

$AC^2 = (24)^2 + (7)^2$

$AC^2 = 576 + 49$

$AC^2 = 625$

$AC = \sqrt{625}$

$AC = 25$ cm (Since length must be positive)

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(i) Values of $\sin$ A and $\cos$ A (with respect to angle A):

The side opposite to angle A is BC.

The side adjacent to angle A is AB.

The hypotenuse is AC.

$\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$

$\cos A = \frac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$

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(ii) Values of $\sin$ C and $\cos$ C (with respect to angle C):

The side opposite to angle C is AB.

The side adjacent to angle C is BC.

The hypotenuse is AC.

$\sin C = \frac{\text{Side opposite to angle C}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$

$\cos C = \frac{\text{Side adjacent to angle C}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$

Given:

In $\triangle$ PQR, $\angle$Q = $90^\circ$, PQ = 12 cm, PR = 13 cm.

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To Find:

$\tan$ P – $\cot$ R.

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Solution:

In the right-angled $\triangle$ PQR, by the Pythagorean theorem, we have:

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

$(13)^2 = (12)^2 + QR^2$

$169 = 144 + QR^2$

$QR^2 = 169 - 144$

$QR^2 = 25$

$QR = \sqrt{25}$

$QR = 5$ cm (Since length must be positive)

Now, we find the values of $\tan$ P and $\cot$ R.

For angle P:

The side opposite to angle P is QR.

The side adjacent to angle P is PQ.

$\tan P = \frac{\text{Side opposite to angle P}}{\text{Side adjacent to angle P}} = \frac{QR}{PQ} = \frac{5}{12}$

For angle R:

The side opposite to angle R is PQ.

The side adjacent to angle R is QR.

$\cot R = \frac{\text{Side adjacent to angle R}}{\text{Side opposite to angle R}} = \frac{QR}{PQ} = \frac{5}{12}$

Now, we calculate $\tan$ P – $\cot$ R:

$\tan P - \cot R = \frac{5}{12} - \frac{5}{12}$

$\tan P - \cot R = 0$

Given:

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

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To Find:

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

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Solution:

We are given $\sin A = \frac{3}{4}$.

In a right-angled triangle, $\sin A$ is defined as the ratio of the side opposite to angle A to the hypotenuse.

Let's consider a right triangle ABC, right-angled at B, with respect to angle A.

$\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC}$

So, $\frac{BC}{AC} = \frac{3}{4}$.

Let $BC = 3k$ and $AC = 4k$, where $k$ is a positive constant.

By the Pythagorean theorem in $\triangle$ ABC:

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

$(4k)^2 = AB^2 + (3k)^2$

$16k^2 = AB^2 + 9k^2$

$AB^2 = 16k^2 - 9k^2$

$AB^2 = 7k^2$

$AB = \sqrt{7k^2} = k\sqrt{7}$ (Since length is positive)

Now, we can find $\cos A$ and $\tan A$:

$\cos A = \frac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \frac{AB}{AC}$

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

$\tan A = \frac{\text{Side opposite to angle A}}{\text{Side adjacent to angle A}} = \frac{BC}{AB}$

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

We can rationalize the denominator for $\tan A$:

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

Thus, $\cos A = \frac{\sqrt{7}}{4}$ and $\tan A = \frac{3\sqrt{7}}{7}$ (or $\frac{3}{\sqrt{7}}$).

Given:

$15 \cot A = 8$

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To Find:

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

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Solution:

We are given the equation $15 \cot A = 8$.

Dividing both sides by 15, we get:

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

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

Let's consider a right triangle ABC, right-angled at B, with respect to angle A.

$\cot A = \frac{\text{Side adjacent to angle A}}{\text{Side opposite to angle A}} = \frac{AB}{BC}$

So, $\frac{AB}{BC} = \frac{8}{15}$.

Let $AB = 8k$ and $BC = 15k$, where $k$ is a positive constant.

By the Pythagorean theorem in $\triangle$ ABC:

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

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

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

$AC^2 = 289k^2$

$AC = \sqrt{289k^2} = 17k$ (Since length must be positive)

Now, we can find $\sin A$ and $\sec A$:

$\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC}$

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

$\sec A = \frac{\text{Hypotenuse}}{\text{Side adjacent to angle A}} = \frac{AC}{AB}$

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

Thus, $\sin A = \frac{15}{17}$ and $\sec A = \frac{17}{8}$.

Given:

$\sec \theta = \frac{13}{12}$

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To Find:

All other trigonometric ratios.

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Solution:

We are given $\sec \theta = \frac{13}{12}$.

In a right-angled triangle, the trigonometric ratio $\sec \theta$ is defined as the ratio of the hypotenuse to the side adjacent to angle $\theta$.

Let's consider a right triangle ABC, right-angled at B, with respect to angle $\theta$ at A.

$\sec \theta = \frac{\text{Hypotenuse}}{\text{Side adjacent to angle }\theta} = \frac{AC}{AB}$

So, $\frac{AC}{AB} = \frac{13}{12}$.

Let $AC = 13k$ and $AB = 12k$, where $k$ is a positive constant.

By the Pythagorean theorem in $\triangle$ ABC:

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

$(13k)^2 = (12k)^2 + BC^2$

$169k^2 = 144k^2 + BC^2$

$BC^2 = 169k^2 - 144k^2$

$BC^2 = 25k^2$

$BC = \sqrt{25k^2} = 5k$ (Since length must be positive)

Now, we can find the other trigonometric ratios:

$\sin \theta = \frac{\text{Side opposite to angle }\theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}$

$\cos \theta = \frac{\text{Side adjacent to angle }\theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}$

$\tan \theta = \frac{\text{Side opposite to angle }\theta}{\text{Side adjacent to angle }\theta} = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}$

$\text{cosec } \theta = \frac{\text{Hypotenuse}}{\text{Side opposite to angle }\theta} = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5}$

$\cot \theta = \frac{\text{Side adjacent to angle }\theta}{\text{Side opposite to angle }\theta} = \frac{AB}{BC} = \frac{12k}{5k} = \frac{12}{5}$

The other trigonometric ratios are:

$\sin \theta = \frac{5}{13}$

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

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

$\text{cosec } \theta = \frac{13}{5}$

$\cot \theta = \frac{12}{5}$

Given:

$\angle$A and $\angle$B are acute angles.

$\cos$ A = $\cos$ B.

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To Show:

$\angle$A = $\angle$B.

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Solution:

Consider a right-angled triangle ABC, where $\angle$C is the right angle ($90^\circ$).

In this triangle, A and B are the acute angles.

By the definition of the cosine ratio:

For angle A, $\cos$ A = $\frac{\text{Side adjacent to }\angle\text{A}}{\text{Hypotenuse}} = \frac{AC}{AB}$

For angle B, $\cos$ B = $\frac{\text{Side adjacent to }\angle\text{B}}{\text{Hypotenuse}} = \frac{BC}{AB}$

We are given that $\cos$ A = $\cos$ B.

Since AB is the hypotenuse and cannot be zero, we can multiply both sides by AB:

$AC = BC$

In $\triangle$ ABC, if two sides are equal (AC = BC), then the angles opposite to these sides are also equal.

The angle opposite to side AC is $\angle$B.

The angle opposite to side BC is $\angle$A.

Therefore,

Since $\angle$A and $\angle$B are given as acute angles, and we have shown they are equal, the condition holds.

Hence Shown.

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Alternate Solution (Using Trigonometric Identity/Inverse):

Given $\cos A = \cos B$, where A and B are acute angles.

Since the cosine function is a one-to-one function for acute angles (in the domain $0^\circ \leq \theta \leq 90^\circ$), if $\cos A = \cos B$, then the angles A and B must be equal.

Using the inverse cosine function:

$A = \cos^{-1}(\cos A)$

$B = \cos^{-1}(\cos B)$

Since $\cos A = \cos B$, we have $\cos^{-1}(\cos A) = \cos^{-1}(\cos B)$.

Therefore, $A = B$.

Thus, $\angle$A = $\angle$B.

Hence Shown.

Given:

$\cot \theta = \frac{7}{8}$

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To Evaluate:

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

(ii) $\cot^2 \theta$

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Solution:

(i) Evaluate $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$:

We can simplify the expression using the algebraic identity $(a+b)(a-b) = a^2 - b^2$ in both the numerator and the denominator.

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

Denominator: $(1 + \cos \theta)(1 - \cos \theta) = 1^2 - \cos^2 \theta = 1 - \cos^2 \theta$

So the expression becomes:

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

Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we have:

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

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

Substitute these into the expression:

$\frac{\cos^2 \theta}{\sin^2 \theta}$

We know that $\frac{\cos \theta}{\sin \theta} = \cot \theta$. Therefore, $\frac{\cos^2 \theta}{\sin^2 \theta} = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \cot^2 \theta$.

The expression simplifies to $\cot^2 \theta$.

We are given $\cot \theta = \frac{7}{8}$.

So, $\cot^2 \theta = \left(\frac{7}{8}\right)^2 = \frac{7^2}{8^2} = \frac{49}{64}$.

Thus, $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)} = \frac{49}{64}$.

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(ii) Evaluate $\cot^2 \theta$:

We are given $\cot \theta = \frac{7}{8}$.

$\cot^2 \theta = (\cot \theta)^2 = \left(\frac{7}{8}\right)^2$

$\cot^2 \theta = \frac{7^2}{8^2} = \frac{49}{64}$

The value of $\cot^2 \theta$ is $\frac{49}{64}$.

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Alternate Method for (i) (using a triangle):

Given $\cot \theta = \frac{7}{8}$.

In a right triangle, let the side adjacent to $\theta$ be $7k$ and the side opposite to $\theta$ be $8k$.

Using the Pythagorean theorem, the hypotenuse is $\sqrt{(7k)^2 + (8k)^2} = \sqrt{49k^2 + 64k^2} = \sqrt{113k^2} = k\sqrt{113}$.

Then, $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{8k}{k\sqrt{113}} = \frac{8}{\sqrt{113}}$.

And $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{7k}{k\sqrt{113}} = \frac{7}{\sqrt{113}}$.

Now substitute these values into the expression $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$:

$\frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \frac{1 - \left(\frac{8}{\sqrt{113}}\right)^2}{1 - \left(\frac{7}{\sqrt{113}}\right)^2}$

$= \frac{1 - \frac{64}{113}}{1 - \frac{49}{113}}$

$= \frac{\frac{113 - 64}{113}}{\frac{113 - 49}{113}}$

$= \frac{\frac{49}{113}}{\frac{64}{113}}$

$= \frac{49}{113} \times \frac{113}{64}$

$= \frac{\cancel{113}^{1}}{\cancel{113}_{1}} \times \frac{49}{64} = \frac{49}{64}$

This confirms the result obtained using trigonometric identities.

Given:

$3 \cot A = 4$

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To Check:

Whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ or not.

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Solution:

From the given equation, $3 \cot A = 4$, we have:

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

We know that $\tan A = \frac{1}{\cot A}$.

So, $\tan A = \frac{1}{4/3} = \frac{3}{4}$.

Now, let's construct a right-angled triangle ABC, right-angled at B, with respect to angle A.

$\tan A = \frac{\text{Side opposite to A}}{\text{Side adjacent to A}} = \frac{BC}{AB}$

So, $\frac{BC}{AB} = \frac{3}{4}$.

Let $BC = 3k$ and $AB = 4k$, where $k$ is a positive constant.

By the Pythagorean theorem, the hypotenuse AC is:

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

$AC^2 = (4k)^2 + (3k)^2$

$AC^2 = 16k^2 + 9k^2$

$AC^2 = 25k^2$

$AC = \sqrt{25k^2} = 5k$ (Since length is positive)

Now we find the values of $\sin A$ and $\cos A$:

$\sin A = \frac{\text{Side opposite to A}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5}$

$\cos A = \frac{\text{Side adjacent to A}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}$

Now, let's evaluate the left side (LHS) of the given equation:

LHS = $\frac{1 - \tan^2 A}{1 + \tan^2 A}$

Substitute the value of $\tan A = \frac{3}{4}$:

LHS = $\frac{1 - (\frac{3}{4})^2}{1 + (\frac{3}{4})^2} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}}$

LHS = $\frac{\frac{16 - 9}{16}}{\frac{16 + 9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}}$

LHS = $\frac{7}{16} \times \frac{16}{25} = \frac{7}{\cancel{16}} \times \frac{\cancel{16}}{25} = \frac{7}{25}$

Now, let's evaluate the right side (RHS) of the given equation:

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

Substitute the values of $\cos A = \frac{4}{5}$ and $\sin A = \frac{3}{5}$:

RHS = $\left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2$

RHS = $\frac{16}{25} - \frac{9}{25}$

RHS = $\frac{16 - 9}{25} = \frac{7}{25}$

Comparing the LHS and RHS, we see that:

Since LHS = RHS, the equality holds true.

Therefore, $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$.

Given:

In $\triangle$ ABC, right-angled at B, $\tan A = \frac{1}{\sqrt{3}}$.

---

To Find:

(i) $\sin$ A $\cos$ C + $\cos$ A $\sin$ C

(ii) $\cos$ A $\cos$ C – $\sin$ A $\sin$ C

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Solution:

We are given $\tan A = \frac{1}{\sqrt{3}}$.

In a right-angled triangle ABC, right-angled at B:

$\tan A = \frac{\text{Side opposite to A}}{\text{Side adjacent to A}} = \frac{BC}{AB}$

So, $\frac{BC}{AB} = \frac{1}{\sqrt{3}}$.

Let $BC = k$ and $AB = k\sqrt{3}$, where $k$ is a positive constant.

By the Pythagorean theorem, the hypotenuse AC is:

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

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

$AC^2 = 3k^2 + k^2$

$AC^2 = 4k^2$

$AC = \sqrt{4k^2} = 2k$ (Since length is positive)

Now we find the trigonometric ratios for angles A and C.

For angle A:

$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$

$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{k\sqrt{3}}{2k} = \frac{\sqrt{3}}{2}$

For angle C:

The side opposite to angle C is AB.

The side adjacent to angle C is BC.

$\sin C = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{k\sqrt{3}}{2k} = \frac{\sqrt{3}}{2}$

$\cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$

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(i) Evaluate $\sin$ A $\cos$ C + $\cos$ A $\sin$ C:

Substitute the values we found:

$\sin A \cos C + \cos A \sin C = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right)$

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

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

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

So, $\sin A \cos C + \cos A \sin C = 1$.

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(ii) Evaluate $\cos$ A $\cos$ C – $\sin$ A $\sin$ C:

Substitute the values we found:

$\cos A \cos C – \sin A \sin C = \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{1}{2}\right) - \left(\frac{1}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right)$

$= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}$

$= 0$

So, $\cos A \cos C – \sin A \sin C = 0$.

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Alternate Solution (using angles):

Given $\tan A = \frac{1}{\sqrt{3}}$ in a right triangle at B.

We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$. Since A is an acute angle, $A = 30^\circ$.

In a right triangle ABC, $\angle A + \angle B + \angle C = 180^\circ$.

$30^\circ + 90^\circ + \angle C = 180^\circ$

$120^\circ + \angle C = 180^\circ$

$\angle C = 180^\circ - 120^\circ = 60^\circ$.

So, $A = 30^\circ$ and $C = 60^\circ$.

Now we find the values of the trigonometric ratios for these angles:

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

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

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

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

(i) Evaluate $\sin$ A $\cos$ C + $\cos$ A $\sin$ C:

$\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right)$

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

(ii) Evaluate $\cos$ A $\cos$ C – $\sin$ A $\sin$ C:

$\cos 30^\circ \cos 60^\circ – \sin 30^\circ \sin 60^\circ = \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{1}{2}\right) - \left(\frac{1}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right)$

$= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0$

Given:

In $\triangle$ PQR, $\angle$Q = $90^\circ$, PQ = 5 cm, PR + QR = 25 cm.

---

To Find:

The values of $\sin$ P, $\cos$ P, and $\tan$ P.

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Solution:

In the right-angled $\triangle$ PQR, by the Pythagorean theorem, we have:

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

We are given PQ = 5 cm and PR + QR = 25 cm.

From the relation PR + QR = 25, we can express PR as:

PR = 25 – QR

Substitute this expression for PR into the Pythagorean theorem:

$(25 – QR)^2 = (5)^2 + QR^2$

Expand the left side using the identity $(a-b)^2 = a^2 - 2ab + b^2$:

$(25)^2 – 2(25)(QR) + QR^2 = 25 + QR^2$

$625 – 50QR + QR^2 = 25 + QR^2$

Subtract $QR^2$ from both sides of the equation:

$625 – 50QR = 25$

Rearrange the terms to solve for QR:

$625 – 25 = 50QR$

$600 = 50QR$

$QR = \frac{600}{50}$

$QR = 12$ cm

Now find the value of PR using the relation PR = 25 – QR:

PR = 25 – 12

PR = 13 cm

So, the lengths of the sides of $\triangle$ PQR are PQ = 5 cm, QR = 12 cm, and PR = 13 cm.

Now, we determine the trigonometric ratios for angle P:

The side opposite to angle P is QR.

The side adjacent to angle P is PQ.

The hypotenuse is PR.

$\sin P = \frac{\text{Side opposite to angle P}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{12}{13}$

$\cos P = \frac{\text{Side adjacent to angle P}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{5}{13}$

$\tan P = \frac{\text{Side opposite to angle P}}{\text{Side adjacent to angle P}} = \frac{QR}{PQ} = \frac{12}{5}$

The values are $\sin P = \frac{12}{13}$, $\cos P = \frac{5}{13}$, and $\tan P = \frac{12}{5}$.

(i) The value of tan A is always less than 1.

False.

Justification:

In a right-angled triangle, $\tan A = \frac{\text{Side opposite to angle A}}{\text{Side adjacent to angle A}}$.

The length of the side opposite to angle A can be greater than, equal to, or less than the length of the side adjacent to angle A, depending on the value of angle A.

For example, if $A = 45^\circ$, the opposite side equals the adjacent side, so $\tan 45^\circ = 1$.

If $A > 45^\circ$ (and $A < 90^\circ$), the side opposite to A is greater than the side adjacent to A, so $\tan A > 1$. For instance, $\tan 60^\circ = \sqrt{3} \approx 1.732$, which is greater than 1.

Therefore, the value of $\tan A$ is not always less than 1.

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(ii) sec A = $\frac{12}{5}$ for some value of angle A.

True.

Justification:

In a right-angled triangle, $\sec A = \frac{\text{Hypotenuse}}{\text{Side adjacent to angle A}}$.

The hypotenuse is always the longest side in a right-angled triangle (except for the degenerate case of a right angle with zero area, where sides can be zero, but for an angle in a triangle, sides must be positive). Thus, the ratio of the hypotenuse to the adjacent side must be greater than 1.

Given $\sec A = \frac{12}{5} = 2.4$. Since $2.4 > 1$, this value is a valid value for $\sec A$ for some acute angle A.

We can form a right triangle with adjacent side 5 units and hypotenuse 12 units. By the Pythagorean theorem, the opposite side would be $\sqrt{12^2 - 5^2} = \sqrt{144 - 25} = \sqrt{119}$ units. Since $\sqrt{119}$ is a real positive number, such a triangle exists, and $\sec A = \frac{12}{5}$ is possible.

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(iii) cos A is the abbreviation used for the cosecant of angle A.

False.

Justification:

$\cos A$ is the abbreviation used for the cosine of angle A.

The abbreviation used for the cosecant of angle A is $\text{cosec } A$ (or $\text{csc } A$).

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(iv) cot A is the product of cot and A.

False.

Justification:

'cot' is part of the trigonometric ratio name 'cotangent'. $\cot A$ represents the cotangent of the angle A. It is not a product of 'cot' and the angle A. The expression $\cot A$ is a single entity representing the ratio of the adjacent side to the opposite side for the angle A in a right triangle.

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(v) sin θ = $\frac{4}{3}$ for some angle θ.

False.

Justification:

In a right-angled triangle, $\sin \theta = \frac{\text{Side opposite to angle }\theta}{\text{Hypotenuse}}$.

The length of the side opposite to any angle in a right triangle is always less than or equal to the length of the hypotenuse. Therefore, the ratio $\frac{\text{Opposite}}{\text{Hypotenuse}}$ must be less than or equal to 1.

Given $\sin \theta = \frac{4}{3} \approx 1.33$. Since $1.33 > 1$, this value is not possible for $\sin \theta$ for any real angle $\theta$. The range of the sine function is $[-1, 1]$. For angles in a triangle (which are between $0^\circ$ and $180^\circ$), the value of sine is between 0 and 1 (inclusive).

Given:

In $\triangle$ ABC, $\angle$B = $90^\circ$, AB = 5 cm, and $\angle$ACB = $30^\circ$.

---

To Determine:

The lengths of the sides BC and AC.

---

Solution:

We are given a right-angled triangle ABC, with $\angle$B = $90^\circ$, AB = 5 cm, and $\angle$C = $30^\circ$.

We can use trigonometric ratios with respect to the given angle $\angle$C ($30^\circ$) to find the unknown sides.

To find BC (the side adjacent to $\angle$C):

We know the length of the side opposite to $\angle$C (AB = 5 cm). The ratio involving the opposite and adjacent sides is tangent.

$\tan C = \frac{\text{Side opposite to } \angle \text{C}}{\text{Side adjacent to } \angle \text{C}} = \frac{AB}{BC}$

Substitute the given values:

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

We know that the value of $\tan 30^\circ$ is $\frac{1}{\sqrt{3}}$.

Cross-multiply to solve for BC:

$BC \times 1 = 5 \times \sqrt{3}$

$BC = 5\sqrt{3}$ cm

To find AC (the hypotenuse):

We can use the sine ratio, which involves the opposite side and the hypotenuse.

$\sin C = \frac{\text{Side opposite to } \angle \text{C}}{\text{Hypotenuse}} = \frac{AB}{AC}$

Substitute the given values:

$\sin 30^\circ = \frac{5}{AC}$

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

Cross-multiply to solve for AC:

$AC \times 1 = 5 \times 2$

$AC = 10$ cm

Alternatively, once BC is found, we could use the cosine ratio for $\angle$C, or the Pythagorean theorem to find AC.

Using $\cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC}$:

$\cos 30^\circ = \frac{5\sqrt{3}}{AC}$

$\frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{AC}$

$AC \times \sqrt{3} = 2 \times 5\sqrt{3}$

$AC = \frac{10\sqrt{3}}{\sqrt{3}} = 10$ cm.

Using the Pythagorean theorem $AC^2 = AB^2 + BC^2$:

$AC^2 = (5)^2 + (5\sqrt{3})^2$

$AC^2 = 25 + (25 \times 3)$

$AC^2 = 25 + 75$

$AC^2 = 100$

$AC = \sqrt{100} = 10$ cm.

The lengths of the sides are BC = $5\sqrt{3}$ cm and AC = 10 cm.

Given:

In $\triangle$ PQR, $\angle$Q = $90^\circ$, PQ = 3 cm, and PR = 6 cm.

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To Determine:

$\angle$QPR and $\angle$PRQ.

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Solution:

In the right-angled $\triangle$ PQR, we have PQ = 3 cm and PR = 6 cm. The hypotenuse is PR.

Consider angle P ($\angle$QPR):

The side adjacent to angle P is PQ.

The hypotenuse is PR.

We can use the cosine ratio:

$\cos P = \frac{\text{Side adjacent to P}}{\text{Hypotenuse}} = \frac{PQ}{PR}$

$\cos P = \frac{3}{6} = \frac{1}{2}$

We know that $\cos 60^\circ = \frac{1}{2}$.

Since P is an acute angle in a right triangle, comparing the values, we get:

$P = 60^\circ$

So, $\angle$QPR = $60^\circ$.

Consider angle R ($\angle$PRQ):

The side opposite to angle R is PQ.

The hypotenuse is PR.

We can use the sine ratio:

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

$\sin R = \frac{3}{6} = \frac{1}{2}$

We know that $\sin 30^\circ = \frac{1}{2}$.

Since R is an acute angle in a right triangle, comparing the values, we get:

$R = 30^\circ$

So, $\angle$PRQ = $30^\circ$.

Alternatively, once we found $\angle$P = $60^\circ$, we can find $\angle$R using the angle sum property of a triangle:

In $\triangle$ PQR, $\angle$P + $\angle$Q + $\angle$R = $180^\circ$

$60^\circ + 90^\circ + \angle$R = $180^\circ$

$150^\circ + \angle$R = $180^\circ$

$\angle$R = $180^\circ - 150^\circ = 30^\circ$.

So, $\angle$PRQ = $30^\circ$.

The angles are $\angle$QPR = $60^\circ$ and $\angle$PRQ = $30^\circ$.

Given:

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

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

$0^\circ < A + B \leq 90^\circ$

$A > B$

---

To Find:

The values of angles A and B.

---

Solution:

We are given the equation $\sin (A - B) = \frac{1}{2}$.

We know that $\sin 30^\circ = \frac{1}{2}$.

Since the sine function is one-to-one for angles in the range typically considered in these problems (especially given A > B and the context of acute angles), we can equate the arguments:

We are also given the equation $\cos (A + B) = \frac{1}{2}$.

We know that $\cos 60^\circ = \frac{1}{2}$.

The condition $0^\circ < A + B \leq 90^\circ$ tells us that A + B is an acute angle or a right angle in the first quadrant boundary. Since $\cos (A+B) = \frac{1}{2}$ (which is positive), A + B must be in the first quadrant.

Therefore, we can equate the arguments:

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

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

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

Add equation (i) and equation (ii):

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

$2A = 90^\circ$

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

$A = 45^\circ$

Substitute the value of A into equation (ii):

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

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

$B = 15^\circ$

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

$A = 45^\circ$, $B = 15^\circ$.

$A > B$: $45^\circ > 15^\circ$ (True)

$0^\circ < A + B \leq 90^\circ$: $A + B = 45^\circ + 15^\circ = 60^\circ$. $0^\circ < 60^\circ \leq 90^\circ$ (True)

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

(i) Evaluate $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$:

We use the standard trigonometric values:

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

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

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

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

Substitute these values into the expression:

$\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)$

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

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

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

The value is 1.

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(ii) Evaluate $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$:

We use the standard trigonometric values:

$\tan 45^\circ = 1$

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

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

Substitute these values into the expression:

$2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ = 2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2$

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

$= 2 + 0 = 2$

The value is 2.

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(iii) Evaluate $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$:

We use the standard trigonometric values:

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

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

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

Substitute these values into the expression:

$\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ} = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$

Find a common denominator in the denominator:

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

Invert the denominator and multiply:

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

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

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

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $2\sqrt{2} - 2\sqrt{6}$ or $2(\sqrt{2} - \sqrt{6})$. It's simpler to factor out 2 first.

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

Now multiply by $\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$ (conjugate of $\sqrt{2} + \sqrt{6}$ after swapping terms for easier calculation of difference of squares):

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

$= \frac{\sqrt{18}-\sqrt{6}}{2((\sqrt{6})^2-(\sqrt{2})^2)}$

$= \frac{\sqrt{9 \times 2}-\sqrt{6}}{2(6-2)}$

$= \frac{3\sqrt{2}-\sqrt{6}}{2(4)}$

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

The value is $\frac{3\sqrt{2}-\sqrt{6}}{8}$.

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(iv) Evaluate $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$:

We use the standard trigonometric values:

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

$\tan 45^\circ = 1$

$\text{cosec } 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$

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

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

$\cot 45^\circ = 1$

Substitute these values into the expression:

Numerator = $\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ = \frac{1}{2} + 1 - \frac{2}{\sqrt{3}}$

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

Find a common denominator: $\frac{3\sqrt{3} - 4}{2\sqrt{3}}$

Denominator = $\sec 30^\circ + \cos 60^\circ + \cot 45^\circ = \frac{2}{\sqrt{3}} + \frac{1}{2} + 1$

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

Find a common denominator: $\frac{4 + 3\sqrt{3}}{2\sqrt{3}}$

Now divide the numerator by the denominator:

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

$= \frac{3\sqrt{3} - 4}{\cancel{2\sqrt{3}}} \times \frac{\cancel{2\sqrt{3}}}{3\sqrt{3} + 4}$

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

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $3\sqrt{3} - 4$:

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

Use $(a-b)^2 = a^2 - 2ab + b^2$ 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 = $(3\sqrt{3})^2 - (4)^2 = (9 \times 3) - 16 = 27 - 16 = 11$

The expression is equal to $\frac{43 - 24\sqrt{3}}{11}$.

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(v) Evaluate $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$:

We use the standard trigonometric values:

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

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

$\tan 45^\circ = 1$

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

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

Substitute these values into the expression:

Numerator = $5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ = 5\left(\frac{1}{2}\right)^2 + 4\left(\frac{2}{\sqrt{3}}\right)^2 - (1)^2$

$= 5\left(\frac{1}{4}\right) + 4\left(\frac{4}{3}\right) - 1$

$= \frac{5}{4} + \frac{16}{3} - 1$

Find a common denominator (12):

$= \frac{5 \times 3}{4 \times 3} + \frac{16 \times 4}{3 \times 4} - \frac{1 \times 12}{12}$

$= \frac{15}{12} + \frac{64}{12} - \frac{12}{12}$

$= \frac{15 + 64 - 12}{12} = \frac{79 - 12}{12} = \frac{67}{12}$

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

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

Alternatively, using the identity $\sin^2 \theta + \cos^2 \theta = 1$, the denominator is simply 1.

Now divide the numerator by the denominator:

$\frac{\frac{67}{12}}{1} = \frac{67}{12}$

The value is $\frac{67}{12}$.

(i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$:

We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

Substitute this value into the expression:

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

$= \frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$

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

$= \frac{\cancel{2}^1}{\sqrt{3}} \times \frac{3}{\cancel{4}^2} = \frac{3}{2\sqrt{3}}$

Rationalize the denominator:

$= \frac{3}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{2(3)} = \frac{\cancel{3}\sqrt{3}}{2\cancel{3}} = \frac{\sqrt{3}}{2}$

Now, we compare this value with the given options:

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

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

(C) $\tan 60^\circ = \sqrt{3}$

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

The value of the expression is equal to $\sin 60^\circ$.

The correct option is (A).

Justification: Calculation shows the expression evaluates to $\frac{\sqrt{3}}{2}$, which is the value of $\sin 60^\circ$.

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(ii) $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}$:

We know that $\tan 45^\circ = 1$.

Substitute this value into the expression:

$\frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$

Now, we compare this value with the given options:

(A) $\tan 90^\circ$ (Undefined)

(B) 1

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

(D) 0

The value of the expression is 0.

The correct option is (D).

Justification: Calculation shows the expression evaluates to 0.

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(iii) $\sin 2A = 2 \sin A$ is true when A =:

We need to check which value of A satisfies the equation $\sin 2A = 2 \sin A$.

(A) If $A = 0^\circ$:

LHS = $\sin (2 \times 0^\circ) = \sin 0^\circ = 0$

RHS = $2 \sin 0^\circ = 2 \times 0 = 0$

LHS = RHS. So, A = $0^\circ$ is a solution.

(B) If $A = 30^\circ$:

LHS = $\sin (2 \times 30^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$

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

LHS $\neq$ RHS.

(C) If $A = 45^\circ$:

LHS = $\sin (2 \times 45^\circ) = \sin 90^\circ = 1$

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

LHS $\neq$ RHS.

(D) If $A = 60^\circ$:

LHS = $\sin (2 \times 60^\circ) = \sin 120^\circ$. $\sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$

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

LHS $\neq$ RHS.

The equation $\sin 2A = 2 \sin A$ is true only when $A = 0^\circ$ (or other values not typically covered in basic trigonometry). Based on the options provided, only A = $0^\circ$ satisfies the condition.

The correct option is (A).

Justification: We checked each option and found that only A = $0^\circ$ satisfies the equation $\sin 2A = 2 \sin A$. (Using the double angle identity $\sin 2A = 2 \sin A \cos A$, the equation becomes $2 \sin A \cos A = 2 \sin A$, which simplifies to $2 \sin A \cos A - 2 \sin A = 0$, or $2 \sin A (\cos A - 1) = 0$. This is true if $\sin A = 0$ or $\cos A = 1$. Both occur when $A = 0^\circ$.)

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(iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ}$:

We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

Substitute this value into the expression:

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

$= \frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$

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

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

Rationalize the denominator:

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

Now, we compare this value with the given options:

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

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

(C) $\tan 60^\circ = \sqrt{3}$

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

The value of the expression is equal to $\tan 60^\circ$.

The correct option is (C).

Justification: Calculation shows the expression evaluates to $\sqrt{3}$, which is the value of $\tan 60^\circ$. (This expression is also the formula for $\tan 2A$, so $\tan(2 \times 30^\circ) = \tan 60^\circ$).

Given:

$\tan (A + B) = \sqrt{3}$

$\tan (A - B) = \frac{1}{\sqrt{3}}$

$0^\circ < A + B \leq 90^\circ$

$A > B$

---

To Find:

The values of angles A and B.

---

Solution:

We are given the equation $\tan (A + B) = \sqrt{3}$.

We know that $\tan 60^\circ = \sqrt{3}$.

Given that $0^\circ < A + B \leq 90^\circ$, and the tangent is positive, A + B must be in the first quadrant. The unique angle in this range whose tangent is $\sqrt{3}$ is $60^\circ$.

Therefore, we can equate the arguments:

We are also given the equation $\tan (A - B) = \frac{1}{\sqrt{3}}$.

We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

Given that $A > B$, the difference $A - B$ must be positive. Also, since $A+B \leq 90^\circ$ and $B > 0$ (otherwise $A=A+B$ and $\tan A = \sqrt{3}$ would mean $A=60^\circ$, then $B=0$, and $\tan(60-0) = \tan 60 = \sqrt{3} \neq 1/\sqrt{3}$), it implies $A < 90^\circ$ and $B$ is also an acute angle, making $A-B$ also likely acute. The unique angle whose tangent is $\frac{1}{\sqrt{3}}$ is $30^\circ$.

Therefore, we can equate the arguments:

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

From equation (i): $A + B = 60^\circ$

From equation (ii): $A - B = 30^\circ$

Add equation (i) and equation (ii):

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

$2A = 90^\circ$

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

A = $45^\circ$

Substitute the value of A = $45^\circ$ into equation (i):

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

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

B = $15^\circ$

Let's verify if these values satisfy the given conditions:

$A = 45^\circ$ and $B = 15^\circ$.

$A > B$: $45^\circ > 15^\circ$. This condition is satisfied.

$0^\circ < A + B \leq 90^\circ$: $A + B = 45^\circ + 15^\circ = 60^\circ$. $0^\circ < 60^\circ \leq 90^\circ$. This condition is satisfied.

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

(i) $\sin (A + B) = \sin A + \sin B$.

False.

Justification:

Consider A = $30^\circ$ and B = $60^\circ$.

LHS = $\sin (A + B) = \sin (30^\circ + 60^\circ) = \sin 90^\circ = 1$.

RHS = $\sin A + \sin B = \sin 30^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}$.

Since $1 \neq \frac{1+\sqrt{3}}{2}$, the statement $\sin (A + B) = \sin A + \sin B$ is generally false.

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(ii) The value of $\sin \theta$ increases as $\theta$ increases.

True (for $0^\circ \leq \theta \leq 90^\circ$).

Justification:

Consider the values of $\sin \theta$ for increasing $\theta$ in the range from $0^\circ$ to $90^\circ$ (which is the typical range for angles in a right triangle as considered in this chapter):

$\sin 0^\circ = 0$

$\sin 30^\circ = 0.5$

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

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

$\sin 90^\circ = 1$

As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\sin \theta$ monotonically increases from 0 to 1. Therefore, the statement is true for this range of $\theta$.

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(iii) The value of $\cos \theta$ increases as $\theta$ increases.

False.

Justification:

Consider the values of $\cos \theta$ for increasing $\theta$ in the range from $0^\circ$ to $90^\circ$:

$\cos 0^\circ = 1$

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

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

$\cos 60^\circ = 0.5$

$\cos 90^\circ = 0$

As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\cos \theta$ monotonically decreases from 1 to 0. Therefore, the statement is false for this range of $\theta$.

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(iv) $\sin \theta = \cos \theta$ for all values of $\theta$.

False.

Justification:

The equality $\sin \theta = \cos \theta$ is only true for specific values of $\theta$, such as $\theta = 45^\circ$ (or $45^\circ + n \times 180^\circ$, where $n$ is an integer). It is not true for all values of $\theta$.

For example, if $\theta = 30^\circ$, $\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$. Since $\frac{1}{2} \neq \frac{\sqrt{3}}{2}$, the statement is false.

---

(v) $\cot A$ is not defined for $A = 0^\circ$.

True.

Justification:

The trigonometric ratio $\cot A$ is defined as $\cot A = \frac{\cos A}{\sin A}$.

For $A = 0^\circ$, we have $\cos 0^\circ = 1$ and $\sin 0^\circ = 0$.

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

Division by zero is undefined. Therefore, $\cot A$ is not defined for $A = 0^\circ$.

To Evaluate:

$\frac{\tan 65^\circ}{\cot 25^\circ}$

---

Solution:

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

Notice that $65^\circ + 25^\circ = 90^\circ$. So, $65^\circ$ and $25^\circ$ are complementary angles.

We can write $65^\circ = 90^\circ - 25^\circ$.

So, $\tan 65^\circ = \tan (90^\circ - 25^\circ)$.

Using the identity $\tan (90^\circ - \theta) = \cot \theta$ with $\theta = 25^\circ$, we get:

$\tan 65^\circ = \cot 25^\circ$

Alternatively, we can write $25^\circ = 90^\circ - 65^\circ$.

So, $\cot 25^\circ = \cot (90^\circ - 65^\circ)$.

Using the identity $\cot (90^\circ - \theta) = \tan \theta$ with $\theta = 65^\circ$, we get:

$\cot 25^\circ = \tan 65^\circ$

Now, substitute either of these results back into the expression:

$\frac{\tan 65^\circ}{\cot 25^\circ} = \frac{\cot 25^\circ}{\cot 25^\circ} = 1$ (using $\tan 65^\circ = \cot 25^\circ$)

or

$\frac{\tan 65^\circ}{\cot 25^\circ} = \frac{\tan 65^\circ}{\tan 65^\circ} = 1$ (using $\cot 25^\circ = \tan 65^\circ$)

The value of the expression is 1.

Given:

$\sin 3A = \cos (A – 26^\circ)$

3A is an acute angle ($0^\circ < 3A < 90^\circ$).

---

To Find:

The value of A.

---

Solution:

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

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

Since 3A is an acute angle, we can write $\sin 3A$ as $\cos (90^\circ - 3A)$.

Substitute this into the given equation:

$\cos (90^\circ - 3A) = \cos (A – 26^\circ)$

Since both angles are likely acute (or within the range where cosine is one-to-one or its values correspond uniquely to acute angles in this context), we can equate the arguments:

$90^\circ - 3A = A – 26^\circ$

Rearrange the terms to solve for A:

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

$116^\circ = 4A$

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

$A = 29^\circ$

Let's check if the conditions are satisfied with A = $29^\circ$:

3A = $3 \times 29^\circ = 87^\circ$. This is an acute angle ($0^\circ < 87^\circ < 90^\circ$), so the condition is satisfied.

A – $26^\circ = 29^\circ – 26^\circ = 3^\circ$. This angle is also acute and positive.

With $A = 29^\circ$, the original equation is $\sin 87^\circ = \cos 3^\circ$. This is true because $\sin 87^\circ = \sin (90^\circ - 3^\circ) = \cos 3^\circ$.

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

To Express:

$\cot 85^\circ + \cos 75^\circ$ in terms of trigonometric ratios of angles between $0^\circ$ and $45^\circ$.

---

Solution:

We need to express the given trigonometric ratios of angles $85^\circ$ and $75^\circ$ (which are greater than $45^\circ$) in terms of trigonometric ratios of angles between $0^\circ$ and $45^\circ$.

We use the complementary angle identities:

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

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

For the term $\cot 85^\circ$:

The angle is $85^\circ$. The complementary angle is $90^\circ - 85^\circ = 5^\circ$.

$\cot 85^\circ = \cot (90^\circ - 5^\circ) = \tan 5^\circ$

The angle $5^\circ$ is between $0^\circ$ and $45^\circ$.

For the term $\cos 75^\circ$:

The angle is $75^\circ$. The complementary angle is $90^\circ - 75^\circ = 15^\circ$.

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

The angle $15^\circ$ is between $0^\circ$ and $45^\circ$.

Now, substitute these expressions back into the original sum:

$\cot 85^\circ + \cos 75^\circ = \tan 5^\circ + \sin 15^\circ$

The expression $\cot 85^\circ + \cos 75^\circ$ is expressed in terms of trigonometric ratios of angles $5^\circ$ and $15^\circ$, both of which are between $0^\circ$ and $45^\circ$.

The expression is $\tan 5^\circ + \sin 15^\circ$.

(i) Evaluate $\frac{\sin 18^\circ}{\cos 72^\circ}$:

We notice that $18^\circ + 72^\circ = 90^\circ$. These are complementary angles.

We use the identity $\sin (90^\circ - \theta) = \cos \theta$ or $\cos (90^\circ - \theta) = \sin \theta$.

Let's express the numerator in terms of cosine:

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

Substitute this into the expression:

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

Assuming $\cos 72^\circ \neq 0$, which is true since $0^\circ < 72^\circ < 90^\circ$, we can cancel the terms.

$= 1$

Alternatively, express the denominator in terms of sine:

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

Substitute this into the expression:

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

Assuming $\sin 18^\circ \neq 0$, which is true since $0^\circ < 18^\circ < 90^\circ$, we can cancel the terms.

$= 1$

The value is 1.

---

(ii) Evaluate $\frac{\tan 26^\circ}{\cot 64^\circ}$:

We notice that $26^\circ + 64^\circ = 90^\circ$. These are complementary angles.

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

Let's express the numerator in terms of cotangent:

$\tan 26^\circ = \tan (90^\circ - 64^\circ) = \cot 64^\circ$

Substitute this into the expression:

$\frac{\tan 26^\circ}{\cot 64^\circ} = \frac{\cot 64^\circ}{\cot 64^\circ}$

Assuming $\cot 64^\circ \neq 0$, which is true since $0^\circ < 64^\circ < 90^\circ$, we can cancel the terms.

$= 1$

The value is 1.

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(iii) Evaluate $\cos 48^\circ – \sin 42^\circ$:

We notice that $48^\circ + 42^\circ = 90^\circ$. These are complementary angles.

We use the identity $\cos (90^\circ - \theta) = \sin \theta$ or $\sin (90^\circ - \theta) = \cos \theta$.

Let's express the first term in terms of sine:

$\cos 48^\circ = \cos (90^\circ - 42^\circ) = \sin 42^\circ$

Substitute this into the expression:

$\cos 48^\circ – \sin 42^\circ = \sin 42^\circ – \sin 42^\circ$

$= 0$

Alternatively, express the second term in terms of cosine:

$\sin 42^\circ = \sin (90^\circ - 48^\circ) = \cos 48^\circ$

Substitute this into the expression:

$\cos 48^\circ – \sin 42^\circ = \cos 48^\circ – \cos 48^\circ$

$= 0$

The value is 0.

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(iv) Evaluate $\text{cosec } 31^\circ – \sec 59^\circ$:

We notice that $31^\circ + 59^\circ = 90^\circ$. These are complementary angles.

We use the identity $\text{cosec } (90^\circ - \theta) = \sec \theta$ or $\sec (90^\circ - \theta) = \text{cosec } \theta$.

Let's express the first term in terms of secant:

$\text{cosec } 31^\circ = \text{cosec } (90^\circ - 59^\circ) = \sec 59^\circ$

Substitute this into the expression:

$\text{cosec } 31^\circ – \sec 59^\circ = \sec 59^\circ – \sec 59^\circ$

$= 0$

Alternatively, express the second term in terms of cosecant:

$\sec 59^\circ = \sec (90^\circ - 31^\circ) = \text{cosec } 31^\circ$

Substitute this into the expression:

$\text{cosec } 31^\circ – \sec 59^\circ = \text{cosec } 31^\circ – \text{cosec } 31^\circ$

$= 0$

The value is 0.

(i) Show that $\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ = 1$:

Consider the Left Hand Side (LHS):

LHS = $\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ$

Group the terms that are complementary:

Notice that $48^\circ + 42^\circ = 90^\circ$ and $23^\circ + 67^\circ = 90^\circ$.

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

$\tan 48^\circ = \tan (90^\circ - 42^\circ) = \cot 42^\circ$

$\tan 23^\circ = \tan (90^\circ - 67^\circ) = \cot 67^\circ$

Substitute these into the LHS:

LHS = $(\cot 42^\circ) (\cot 67^\circ) (\tan 42^\circ) (\tan 67^\circ)$

Rearrange the terms:

LHS = $(\cot 42^\circ \tan 42^\circ) (\cot 67^\circ \tan 67^\circ)$

We know that $\cot \theta \tan \theta = 1$ (since $\cot \theta = \frac{1}{\tan \theta}$).

LHS = $(1) \times (1)$

LHS = 1

This is equal to the Right Hand Side (RHS).

Hence Shown.

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(ii) Show that $\cos 38^\circ \cos 52^\circ – \sin 38^\circ \sin 52^\circ = 0$:

Consider the Left Hand Side (LHS):

LHS = $\cos 38^\circ \cos 52^\circ – \sin 38^\circ \sin 52^\circ$

Notice that $38^\circ + 52^\circ = 90^\circ$. These are complementary angles.

We use the identity $\sin (90^\circ - \theta) = \cos \theta$ and $\cos (90^\circ - \theta) = \sin \theta$.

Let's express the terms involving $52^\circ$ in terms of $38^\circ$:

$\cos 52^\circ = \cos (90^\circ - 38^\circ) = \sin 38^\circ$

$\sin 52^\circ = \sin (90^\circ - 38^\circ) = \cos 38^\circ$

Substitute these into the LHS:

LHS = $\cos 38^\circ (\sin 38^\circ) – \sin 38^\circ (\cos 38^\circ)$

LHS = $\cos 38^\circ \sin 38^\circ – \sin 38^\circ \cos 38^\circ$

The two terms are identical but with opposite signs, so they cancel out.

LHS = 0

This is equal to the Right Hand Side (RHS).

Hence Shown.

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Alternate Solution for (ii) (using angle sum identity):

Recall the cosine addition formula: $\cos (A + B) = \cos A \cos B - \sin A \sin B$.

The expression $\cos 38^\circ \cos 52^\circ – \sin 38^\circ \sin 52^\circ$ matches the form of $\cos (A + B)$ with $A = 38^\circ$ and $B = 52^\circ$.

LHS = $\cos (38^\circ + 52^\circ)$

LHS = $\cos 90^\circ$

We know that $\cos 90^\circ = 0$.

LHS = 0

This is equal to the RHS.

Hence Shown.

Given:

$\tan 2A = \cot (A – 18^\circ)$

2A is an acute angle ($0^\circ < 2A < 90^\circ$).

---

To Find:

The value of A.

---

Solution:

We are given the equation $\tan 2A = \cot (A – 18^\circ)$.

We know the trigonometric identity for complementary angles: $\tan \theta = \cot (90^\circ - \theta)$.

Since 2A is an acute angle, we can write $\tan 2A$ as $\cot (90^\circ - 2A)$.

Substitute this into the given equation:

$\cot (90^\circ - 2A) = \cot (A – 18^\circ)$

Since $2A$ is an acute angle, $90^\circ - 2A$ is also an angle between $0^\circ$ and $90^\circ$. For these angles, the cotangent function is one-to-one. Thus, we can equate the arguments of the cotangent function:

$90^\circ - 2A = A – 18^\circ$

Now, we solve this linear equation for A.

Add $2A$ to both sides:

$90^\circ = A – 18^\circ + 2A$

$90^\circ = 3A – 18^\circ$

Add $18^\circ$ to both sides:

$90^\circ + 18^\circ = 3A$

$108^\circ = 3A$

Divide by 3:

$A = \frac{108^\circ}{3}$

$A = 36^\circ$

Let's verify the condition that 2A is an acute angle:

$2A = 2 \times 36^\circ = 72^\circ$. Since $0^\circ < 72^\circ < 90^\circ$, 2A is indeed an acute angle.

Also, $A - 18^\circ = 36^\circ - 18^\circ = 18^\circ$, which is positive.

The value of A is $36^\circ$.

Given:

$\tan A = \cot B$

Angles A and B are acute angles (implied by the context of trigonometric ratios in a right triangle in this section).

---

To Prove:

A + B = $90^\circ$.

---

Solution:

We are given the equation $\tan A = \cot B$.

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

We can rewrite the right side of the given equation using this identity:

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

Substitute this back into the given equation:

Since A and B are acute angles (and thus $90^\circ - B$ is also an angle between $0^\circ$ and $90^\circ$), and the tangent function is one-to-one for angles between $0^\circ$ and $90^\circ$, we can equate the arguments:

$A = 90^\circ - B$

Add B to both sides of the equation:

$A + B = 90^\circ$

Thus, if $\tan A = \cot B$ and A and B are acute angles, then A + B = $90^\circ$.

Hence Proved.

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Alternate Solution:

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

We can use the identity $\tan A = \frac{\sin A}{\cos A}$ and $\cot B = \frac{\cos B}{\sin B}$.

$\frac{\sin A}{\cos A} = \frac{\cos B}{\sin B}$

Cross-multiply:

$\sin A \sin B = \cos A \cos B$

Rearrange the terms:

$\cos A \cos B - \sin A \sin B = 0$

Recognize the left side as the cosine addition formula $\cos (A + B) = \cos A \cos B - \sin A \sin B$:

$\cos (A + B) = 0$

For acute angles A and B, A + B will be in the range $(0^\circ, 180^\circ)$. The angle in this range whose cosine is 0 is $90^\circ$.

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

Hence Proved.

Given:

$\sec 4A = \text{cosec } (A – 20^\circ)$

4A is an acute angle ($0^\circ < 4A < 90^\circ$).

---

To Find:

The value of A.

---

Solution:

We are given the equation $\sec 4A = \text{cosec } (A – 20^\circ)$.

We use the trigonometric identity for complementary angles: $\sec \theta = \text{cosec } (90^\circ - \theta)$ or $\text{cosec } \theta = \sec (90^\circ - \theta)$.

Since 4A is an acute angle, we can write $\sec 4A$ as $\text{cosec } (90^\circ - 4A)$.

Substitute this into the given equation:

$\text{cosec } (90^\circ - 4A) = \text{cosec } (A – 20^\circ)$

Since $4A$ is an acute angle, $90^\circ - 4A$ is also an angle between $0^\circ$ and $90^\circ$. For these angles, the cosecant function is one-to-one. Thus, we can equate the arguments of the cosecant function:

$90^\circ - 4A = A – 20^\circ$

Now, we solve this linear equation for A.

Add $4A$ to both sides:

$90^\circ = A – 20^\circ + 4A$

$90^\circ = 5A – 20^\circ$

Add $20^\circ$ to both sides:

$90^\circ + 20^\circ = 5A$

$110^\circ = 5A$

Divide by 5:

$A = \frac{110^\circ}{5}$

$A = 22^\circ$

Let's verify the condition that 4A is an acute angle:

$4A = 4 \times 22^\circ = 88^\circ$. Since $0^\circ < 88^\circ < 90^\circ$, 4A is indeed an acute angle.

Also, $A - 20^\circ = 22^\circ - 20^\circ = 2^\circ$, which is positive.

The value of A is $22^\circ$.

Given:

A, B, and C are the interior angles of a triangle ABC.

---

To Show:

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

---

Solution:

Since A, B, and C are the interior angles of a triangle, their sum is $180^\circ$.

We need the expression $\frac{B+C}{2}$. Let's isolate B + C from the equation:

B + C = $180^\circ$ - A

Now, divide both sides of the equation by 2:

$\frac{B + C}{2} = \frac{180^\circ - A}{2}$

$\frac{B + C}{2} = \frac{180^\circ}{2} - \frac{A}{2}$

$\frac{B + C}{2} = 90^\circ - \frac{A}{2}$

Now, take the sine of both sides of this equation:

$\sin \left( \frac{B + C}{2} \right) = \sin \left( 90^\circ - \frac{A}{2} \right)$

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

Here, $\theta = \frac{A}{2}$.

So, $\sin \left( 90^\circ - \frac{A}{2} \right) = \cos \frac{A}{2}$.

Substitute this back into the equation:

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

This matches the required identity.

Hence Shown.

To Express:

$\sin 67^\circ + \cos 75^\circ$ in terms of trigonometric ratios of angles between $0^\circ$ and $45^\circ$.

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Solution:

We need to express the given trigonometric ratios of angles $67^\circ$ and $75^\circ$ (which are greater than $45^\circ$) in terms of trigonometric ratios of angles between $0^\circ$ and $45^\circ$.

We use the complementary angle identities:

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

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

For the term $\sin 67^\circ$:

The angle is $67^\circ$. The complementary angle is $90^\circ - 67^\circ = 23^\circ$.

$\sin 67^\circ = \sin (90^\circ - 23^\circ) = \cos 23^\circ$

The angle $23^\circ$ is between $0^\circ$ and $45^\circ$.

For the term $\cos 75^\circ$:

The angle is $75^\circ$. The complementary angle is $90^\circ - 75^\circ = 15^\circ$.

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

The angle $15^\circ$ is between $0^\circ$ and $45^\circ$.

Now, substitute these expressions back into the original sum:

$\sin 67^\circ + \cos 75^\circ = \cos 23^\circ + \sin 15^\circ$

The expression $\sin 67^\circ + \cos 75^\circ$ is expressed in terms of trigonometric ratios of angles $23^\circ$ and $15^\circ$, both of which are between $0^\circ$ and $45^\circ$.

The expression is $\cos 23^\circ + \sin 15^\circ$.

Given:

Trigonometric ratios $\cos A$, $\tan A$, and $\sec A$.

---

To Express:

$\cos A$, $\tan A$, and $\sec A$ in terms of $\sin A$.

---

Solution:

We use fundamental trigonometric identities.

Expressing $\cos A$ in terms of $\sin A$:

We use the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$.

Subtract $\sin^2 A$ from both sides:

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

Take the square root of both sides:

$\cos A = \pm \sqrt{1 - \sin^2 A}$

In problems involving angles in a right triangle, A is typically an acute angle ($0^\circ < A < 90^\circ$). In this range, $\cos A$ is positive. If A could be other angles, the sign would depend on the quadrant of A.

Assuming A is an acute angle:

$\cos A = \sqrt{1 - \sin^2 A}$

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Expressing $\tan A$ in terms of $\sin A$:

We use the definition of tangent in terms of sine and cosine: $\tan A = \frac{\sin A}{\cos A}$.

Now substitute the expression for $\cos A$ in terms of $\sin A$ (assuming A is acute):

$\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}$

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Expressing $\sec A$ in terms of $\sin A$:

We use the reciprocal identity: $\sec A = \frac{1}{\cos A}$.

Now substitute the expression for $\cos A$ in terms of $\sin A$ (assuming A is acute):

$\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}$

Summary of the expressions in terms of $\sin A$ (assuming A is an acute angle):

$\cos A = \sqrt{1 - \sin^2 A}$

$\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}$

$\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}$

Given:

The identity $\sec A (1 – \sin A)(\sec A + \tan A) = 1$.

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To Prove:

$\sec A (1 – \sin A)(\sec A + \tan A) = 1$

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Proof:

Consider the Left Hand Side (LHS) of the identity:

LHS = $\sec A (1 – \sin A)(\sec A + \tan A)$

Express $\sec A$ and $\tan A$ in terms of $\sin A$ and $\cos A$ using the identities $\sec A = \frac{1}{\cos A}$ and $\tan A = \frac{\sin A}{\cos A}$.

Substitute these into the LHS:

LHS = $\frac{1}{\cos A} (1 – \sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$

Combine the terms within the second parenthesis:

LHS = $\frac{1}{\cos A} (1 – \sin A)\left(\frac{1 + \sin A}{\cos A}\right)$

Multiply the terms in the expression:

LHS = $\frac{1 \times (1 – \sin A) \times (1 + \sin A)}{\cos A \times \cos A}$

LHS = $\frac{(1 – \sin A)(1 + \sin A)}{\cos^2 A}$

Apply the algebraic identity $(a-b)(a+b) = a^2 - b^2$ to the numerator:

LHS = $\frac{1^2 – \sin^2 A}{\cos^2 A}$

LHS = $\frac{1 – \sin^2 A}{\cos^2 A}$

Use the fundamental trigonometric identity $\sin^2 A + \cos^2 A = 1$, which implies $1 – \sin^2 A = \cos^2 A$.

Substitute $1 – \sin^2 A$ with $\cos^2 A$ in the numerator:

LHS = $\frac{\cos^2 A}{\cos^2 A}$

Assuming $\cos^2 A \neq 0$ (which is true for values of A where $\sec A$ and $\tan A$ are defined, i.e., $A \neq 90^\circ, 270^\circ, ...$), we can cancel the $\cos^2 A$ terms:

LHS = 1

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

Given:

The identity $\frac{\cot A – \cos A}{\cot A + \cos A} = \frac{\text{cosec } A – 1}{\text{cosec } A + 1}$.

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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:

Consider the Left Hand Side (LHS) of the identity:

LHS = $\frac{\cot A – \cos A}{\cot A + \cos A}$

Express $\cot A$ in terms of $\sin A$ and $\cos A$ using the identity $\cot A = \frac{\cos A}{\sin A}$.

Substitute this into the LHS:

LHS = $\frac{\frac{\cos A}{\sin A} – \cos A}{\frac{\cos A}{\sin A} + \cos A}$

Factor out $\cos A$ from the numerator and the denominator:

LHS = $\frac{\cos A \left(\frac{1}{\sin A} – 1\right)}{\cos A \left(\frac{1}{\sin A} + 1\right)}$

Assuming $\cos A \neq 0$, we can cancel the $\cos A$ terms:

LHS = $\frac{\frac{1}{\sin A} – 1}{\frac{1}{\sin A} + 1}$

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

Substitute this into the expression:

LHS = $\frac{\text{cosec } A – 1}{\text{cosec } A + 1}$

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

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Alternate Proof (Starting from RHS):

Consider the Right Hand Side (RHS) of the identity:

RHS = $\frac{\text{cosec } A – 1}{\text{cosec } A + 1}$

Express $\text{cosec } A$ in terms of $\sin A$ using the identity $\text{cosec } A = \frac{1}{\sin A}$.

Substitute this into the RHS:

RHS = $\frac{\frac{1}{\sin A} – 1}{\frac{1}{\sin A} + 1}$

Multiply the numerator and the denominator by $\sin A$ to clear the fractions:

RHS = $\frac{\left(\frac{1}{\sin A} – 1\right) \times \sin A}{\left(\frac{1}{\sin A} + 1\right) \times \sin A}$

RHS = $\frac{1 – \sin A}{\sin A} \times \frac{\sin A}{1 + \sin A}$

No, that's not correct. It should be distributing the $\sin A$ in numerator and denominator.

RHS = $\frac{\frac{1}{\sin A} \times \sin A – 1 \times \sin A}{\frac{1}{\sin A} \times \sin A + 1 \times \sin A}$

RHS = $\frac{1 – \sin A}{1 + \sin A}$

Now, relate this back to the LHS. This expression $(1-\sin A)/(1+\sin A)$ is not immediately the form of the LHS. Let's go back to the step where we had the LHS in terms of $\sin A$ and $\cos A$.

LHS = $\frac{\frac{\cos A}{\sin A} – \cos A}{\frac{\cos A}{\sin A} + \cos A}$

Multiply the numerator and denominator by $\sin A$:

LHS = $\frac{(\frac{\cos A}{\sin A} – \cos A) \times \sin A}{(\frac{\cos A}{\sin A} + \cos A) \times \sin A}$

LHS = $\frac{\frac{\cos A}{\sin A} \times \sin A – \cos A \times \sin A}{\frac{\cos A}{\sin A} \times \sin A + \cos A \times \sin A}$

LHS = $\frac{\cos A – \cos A \sin A}{\cos A + \cos A \sin A}$

Factor out $\cos A$ from the numerator and denominator:

LHS = $\frac{\cos A (1 – \sin A)}{\cos A (1 + \sin A)}$

Assuming $\cos A \neq 0$, we can cancel $\cos A$:

LHS = $\frac{1 – \sin A}{1 + \sin A}$

Now consider the RHS again:

RHS = $\frac{\text{cosec } A – 1}{\text{cosec } A + 1}$

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

RHS = $\frac{\frac{1}{\sin A} – 1}{\frac{1}{\sin A} + 1}$

Find a common denominator in the numerator and denominator:

RHS = $\frac{\frac{1 - \sin A}{\sin A}}{\frac{1 + \sin A}{\sin A}}$

Invert the denominator and multiply:

RHS = $\frac{1 - \sin A}{\sin A} \times \frac{\sin A}{1 + \sin A}$

Assuming $\sin A \neq 0$, we can cancel $\sin A$:

RHS = $\frac{1 - \sin A}{1 + \sin A}$

Since LHS = $\frac{1 - \sin A}{1 + \sin A}$ and RHS = $\frac{1 - \sin A}{1 + \sin A}$, we have LHS = RHS.

Hence Proved.

The first proof (starting from LHS and converting to $\sin A$ and $\cos A$) was more direct and simpler.

Given:

The identity $\frac{\sin \theta – \cos \theta + 1}{\sin \theta + \cos \theta – 1} = \frac{1}{\sec \theta – \tan \theta}$.

The identity $\sec^2 \theta = 1 + \tan^2 \theta$ is to be used.

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To Prove:

$\frac{\sin \theta – \cos \theta + 1}{\sin \theta + \cos \theta – 1} = \frac{1}{\sec \theta – \tan \theta}$

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Proof:

Consider the Left Hand Side (LHS) of the identity:

LHS = $\frac{\sin \theta – \cos \theta + 1}{\sin \theta + \cos \theta – 1}$

To express the terms in $\tan \theta$ and $\sec \theta$, divide the numerator and the denominator by $\cos \theta$, assuming $\cos \theta \neq 0$.

LHS = $\frac{\frac{\sin \theta}{\cos \theta} – \frac{\cos \theta}{\cos \theta} + \frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\cos \theta} – \frac{1}{\cos \theta}}$

Using the identities $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$:

LHS = $\frac{\tan \theta – 1 + \sec \theta}{\tan \theta + 1 – \sec \theta}$

Rearrange the terms in the numerator:

LHS = $\frac{\tan \theta + \sec \theta – 1}{\tan \theta – \sec \theta + 1}$

We are asked to use the identity $\sec^2 \theta = 1 + \tan^2 \theta$. This identity can be rearranged to $1 = \sec^2 \theta - \tan^2 \theta$.

Using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$, we have $\sec^2 \theta - \tan^2 \theta = (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)$.

So, $1 = (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)$.

Substitute this expression for '1' into the numerator of the LHS:

Numerator = $(\tan \theta + \sec \theta) – 1$

Numerator = $(\tan \theta + \sec \theta) – (\sec^2 \theta - \tan^2 \theta)$

Numerator = $(\tan \theta + \sec \theta) – (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)$

Now, factor out the common term $(\tan \theta + \sec \theta)$ from the numerator:

Numerator = $(\tan \theta + \sec \theta) [1 – (\sec \theta - \tan \theta)]$

Numerator = $(\tan \theta + \sec \theta) [1 – \sec \theta + \tan \theta]$

Numerator = $(\tan \theta + \sec \theta) [\tan \theta – \sec \theta + 1]$

Substitute this back into the LHS expression:

LHS = $\frac{(\tan \theta + \sec \theta) (\tan \theta – \sec \theta + 1)}{\tan \theta – \sec \theta + 1}$

Assuming $\tan \theta – \sec \theta + 1 \neq 0$, we can cancel the common factor from the numerator and the denominator:

LHS = $\tan \theta + \sec \theta$

Now, consider the Right Hand Side (RHS) of the identity:

RHS = $\frac{1}{\sec \theta – \tan \theta}$

To simplify the RHS and relate it to the LHS result, multiply the numerator and denominator by the conjugate of the denominator, which is $\sec \theta + \tan \theta$, assuming $\sec \theta - \tan \theta \neq 0$ and $\sec \theta + \tan \theta \neq 0$.

RHS = $\frac{1}{\sec \theta – \tan \theta} \times \frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$

RHS = $\frac{\sec \theta + \tan \theta}{(\sec \theta – \tan \theta)(\sec \theta + \tan \theta)}$

Apply the difference of squares formula in the denominator:

RHS = $\frac{\sec \theta + \tan \theta}{\sec^2 \theta – \tan^2 \theta}$

Using the given identity $\sec^2 \theta = 1 + \tan^2 \theta$, we have $\sec^2 \theta – \tan^2 \theta = 1$.

Substitute this into the denominator of the RHS:

RHS = $\frac{\sec \theta + \tan \theta}{1}$

RHS = $\sec \theta + \tan \theta$

Since LHS = $\sec \theta + \tan \theta$ and RHS = $\sec \theta + \tan \theta$, we have:

LHS = RHS

The identity is proven for all values of $\theta$ for which $\cos \theta \neq 0$ (so $\tan \theta$ and $\sec \theta$ are defined) and the denominators in the original expression and during simplification are non-zero.

Hence Proved.

Given:

Trigonometric ratios $\sin A$, $\sec A$, and $\tan A$.

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To Express:

$\sin A$, $\sec A$, and $\tan A$ in terms of $\cot A$.

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Solution:

We use fundamental trigonometric identities.

Expressing $\tan A$ in terms of $\cot A$:

We use the reciprocal identity: $\tan A = \frac{1}{\cot A}$.

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Expressing $\sin A$ in terms of $\cot A$:

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

Taking the square root of both sides:

$\text{cosec } A = \pm \sqrt{1 + \cot^2 A}$

In this chapter, we usually consider acute angles, where trigonometric ratios are positive. So, assuming A is an acute angle:

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

We also know the reciprocal identity $\sin A = \frac{1}{\text{cosec } A}$.

Substitute the expression for $\text{cosec } A$:

$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$

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Expressing $\sec A$ in terms of $\cot A$:

We use the identity $\sec^2 A = 1 + \tan^2 A$.

We already expressed $\tan A$ in terms of $\cot A$: $\tan A = \frac{1}{\cot A}$.

Substitute this into the identity for $\sec^2 A$:

$\sec^2 A = 1 + \left(\frac{1}{\cot A}\right)^2$

$\sec^2 A = 1 + \frac{1}{\cot^2 A}$

Find a common denominator on the right side:

$\sec^2 A = \frac{\cot^2 A + 1}{\cot^2 A}$

Taking the square root of both sides, and assuming A is acute (so $\sec A > 0$):

$\sec A = \sqrt{\frac{\cot^2 A + 1}{\cot^2 A}}$

$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\sqrt{\cot^2 A}}$

$\sec A = \frac{\sqrt{1 + \cot^2 A}}{|\cot A|}$

For an acute angle A, $\cot A$ is positive, so $|\cot A| = \cot A$.

$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$

Summary of the expressions in terms of $\cot A$ (assuming A is an acute angle):

$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$

$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$

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

Given:

Angle A.

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To Express:

All other trigonometric ratios of $\angle$ A ($\sin A$, $\cos A$, $\tan A$, $\text{cosec } A$, $\cot A$) in terms of $\sec A$.

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Solution:

We use fundamental trigonometric identities to express each ratio in terms of $\sec A$. We assume that A is an acute angle, which is standard in this context, unless otherwise specified. For acute angles, all trigonometric ratios are positive, allowing us to take the positive square root.

1. Express $\cos A$ in terms of $\sec A$:

We use the reciprocal identity:

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

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2. Express $\tan A$ in terms of $\sec A$:

We use the identity $\sec^2 A = 1 + \tan^2 A$.

Rearrange the identity to solve for $\tan^2 A$:

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

Take the square root of both sides. For acute angle A, $\tan A$ is positive, so we take the positive root:

$\tan A = \sqrt{\sec^2 A - 1}$

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3. Express $\sin A$ in terms of $\sec A$:

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

Rearrange the identity to solve for $\sin^2 A$:

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

Substitute the expression for $\cos A$ in terms of $\sec A$ ($\cos A = \frac{1}{\sec A}$):

$\sin^2 A = 1 - \left(\frac{1}{\sec A}\right)^2 = 1 - \frac{1}{\sec^2 A}$

Find a common denominator on the right side:

$\sin^2 A = \frac{\sec^2 A - 1}{\sec^2 A}$

Take the square root of both sides. For acute angle A, $\sin A$ is positive, so we take the positive root:

$\sin A = \sqrt{\frac{\sec^2 A - 1}{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{\sqrt{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{|\sec A|}$

For acute angle A, $\sec A$ is positive, so $|\sec A| = \sec A$:

$\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$

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4. Express $\text{cosec } A$ in terms of $\sec A$:

We use the reciprocal identity $\text{cosec } A = \frac{1}{\sin A}$.

Substitute the expression for $\sin A$ in terms of $\sec A$:

$\text{cosec } A = \frac{1}{\frac{\sqrt{\sec^2 A - 1}}{\sec A}}$

$\text{cosec } A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$

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5. Express $\cot A$ in terms of $\sec A$:

We use the reciprocal identity $\cot A = \frac{1}{\tan A}$.

Substitute the expression for $\tan A$ in terms of $\sec A$ (assuming $\tan A > 0$):

$\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$

The other trigonometric ratios of $\angle$ A in terms of $\sec A$ (for acute angle A) are:

$\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$

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

$\tan A = \sqrt{\sec^2 A - 1}$

$\text{cosec } A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$

$\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$

(i) Evaluate $\frac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ}$:

Consider the numerator: $\sin^2 63^\circ + \sin^2 27^\circ$.

Notice that $63^\circ + 27^\circ = 90^\circ$. These are complementary angles.

We use the identity $\sin (90^\circ - \theta) = \cos \theta$ or $\cos (90^\circ - \theta) = \sin \theta$.

We can write $\sin 63^\circ = \sin (90^\circ - 27^\circ) = \cos 27^\circ$.

So, $\sin^2 63^\circ = (\sin 63^\circ)^2 = (\cos 27^\circ)^2 = \cos^2 27^\circ$.

Numerator = $\cos^2 27^\circ + \sin^2 27^\circ$.

Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, with $\theta = 27^\circ$, we get:

Numerator = 1.

Consider the denominator: $\cos^2 17^\circ + \cos^2 73^\circ$.

Notice that $17^\circ + 73^\circ = 90^\circ$. These are complementary angles.

We can write $\cos 73^\circ = \cos (90^\circ - 17^\circ) = \sin 17^\circ$.

So, $\cos^2 73^\circ = (\cos 73^\circ)^2 = (\sin 17^\circ)^2 = \sin^2 17^\circ$.

Denominator = $\cos^2 17^\circ + \sin^2 17^\circ$.

Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, with $\theta = 17^\circ$, we get:

Denominator = 1.

Now, evaluate the entire expression:

$\frac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ} = \frac{1}{1} = 1$

The value is 1.

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(ii) Evaluate $\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ$:

Notice that $25^\circ + 65^\circ = 90^\circ$. These are complementary angles.

We use the identity $\sin (90^\circ - \theta) = \cos \theta$ and $\cos (90^\circ - \theta) = \sin \theta$.

We can write $\cos 65^\circ = \cos (90^\circ - 25^\circ) = \sin 25^\circ$.

We can write $\sin 65^\circ = \sin (90^\circ - 25^\circ) = \cos 25^\circ$.

Substitute these into the expression:

$\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ = \sin 25^\circ (\sin 25^\circ) + \cos 25^\circ (\cos 25^\circ)$

$= \sin^2 25^\circ + \cos^2 25^\circ$

Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, with $\theta = 25^\circ$, we get:

$= 1$

Alternatively, using angle sum identity:

Recall the sine addition formula: $\sin (A + B) = \sin A \cos B + \cos A \sin B$.

The expression $\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ$ matches the form of $\sin (A + B)$ with $A = 25^\circ$ and $B = 65^\circ$.

$\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ = \sin (25^\circ + 65^\circ)$

$= \sin 90^\circ$

We know that $\sin 90^\circ = 1$.

The value is 1.

(i) Evaluate $9 \sec^2 A – 9 \tan^2 A$:

Given expression: $9 \sec^2 A – 9 \tan^2 A$

Factor out the common factor 9:

$= 9 (\sec^2 A – \tan^2 A)$

We use the trigonometric identity $\sec^2 A = 1 + \tan^2 A$.

Rearrange the identity: $\sec^2 A – \tan^2 A = 1$.

Substitute this into the expression:

$= 9 (1)$

$= 9$

The value of the expression is 9.

Comparing with the options, the correct option is (B).

The correct option is (B).

Justification: By factoring and using the identity $\sec^2 A – \tan^2 A = 1$, the expression simplifies to 9.

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(ii) Evaluate $(1 + \tan \theta + \sec \theta)(1 + \cot \theta – \text{cosec } \theta)$:

Given expression: $(1 + \tan \theta + \sec \theta)(1 + \cot \theta – \text{cosec } \theta)$

Express $\tan \theta$, $\sec \theta$, $\cot \theta$, and $\text{cosec } \theta$ in terms of $\sin \theta$ and $\cos \theta$:

$\tan \theta = \frac{\sin \theta}{\cos \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, $\cot \theta = \frac{\cos \theta}{\sin \theta}$, $\text{cosec } \theta = \frac{1}{\sin \theta}$.

Substitute these into the expression:

$(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta})(1 + \frac{\cos \theta}{\sin \theta} – \frac{1}{\sin \theta})$

Find a common denominator within each parenthesis:

$\left(\frac{\cos \theta}{\cos \theta} + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right)\left(\frac{\sin \theta}{\sin \theta} + \frac{\cos \theta}{\sin \theta} – \frac{1}{\sin \theta}\right)$

$\left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right)\left(\frac{\sin \theta + \cos \theta – 1}{\sin \theta}\right)$

Multiply the two fractions:

$\frac{(\cos \theta + \sin \theta + 1)(\sin \theta + \cos \theta – 1)}{\cos \theta \sin \theta}$

In the numerator, group $(\cos \theta + \sin \theta)$ together. The numerator is in the form $(a+b)(a-b)$ where $a = (\cos \theta + \sin \theta)$ and $b = 1$.

Numerator = $((\cos \theta + \sin \theta) + 1)((\cos \theta + \sin \theta) – 1)$

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

Expand $(\cos \theta + \sin \theta)^2$ using $(a+b)^2 = a^2 + 2ab + b^2$:

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

Use the identity $\sin^2 \theta + \cos^2 \theta = 1$:

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

Numerator = $1 + 2 \cos \theta \sin \theta - 1 = 2 \cos \theta \sin \theta$

Substitute the numerator back into the expression:

$\frac{2 \cos \theta \sin \theta}{\cos \theta \sin \theta}$

Assuming $\cos \theta \sin \theta \neq 0$, cancel the common term:

$= 2$

The value of the expression is 2.

Comparing with the options, the correct option is (C).

The correct option is (C).

Justification: By converting the terms to $\sin \theta$ and $\cos \theta$ and simplifying using algebraic and trigonometric identities, the expression evaluates to 2.

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(iii) Evaluate $(\sec A + \tan A)(1 – \sin A)$:

Given expression: $(\sec A + \tan A)(1 – \sin A)$

Express $\sec A$ and $\tan A$ in terms of $\sin A$ and $\cos A$:

$\sec A = \frac{1}{\cos A}$, $\tan A = \frac{\sin A}{\cos A}$.

Substitute these into the first parenthesis:

$\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 – \sin A)$

Combine the terms within the first parenthesis:

$\left(\frac{1 + \sin A}{\cos A}\right)(1 – \sin A)$

Multiply the terms:

$\frac{(1 + \sin A)(1 – \sin A)}{\cos A}$

Apply the algebraic identity $(a+b)(a-b) = a^2 - b^2$ to the numerator:

$\frac{1^2 – \sin^2 A}{\cos A} = \frac{1 – \sin^2 A}{\cos A}$

Use the fundamental trigonometric identity $\sin^2 A + \cos^2 A = 1$, which implies $1 – \sin^2 A = \cos^2 A$.

Substitute $1 – \sin^2 A$ with $\cos^2 A$ in the numerator:

$\frac{\cos^2 A}{\cos A}$

Assuming $\cos A \neq 0$, we can cancel one power of $\cos A$:

$\cos A$

The value of the expression is $\cos A$.

Comparing with the options, the correct option is (D).

The correct option is (D).

Justification: By converting the terms to $\sin A$ and $\cos A$ and simplifying using algebraic and trigonometric identities, the expression evaluates to $\cos A$.

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(iv) Evaluate $\frac{1 + \tan^2 A}{1 + \cot^2 A}$:

Given expression: $\frac{1 + \tan^2 A}{1 + \cot^2 A}$

We use the trigonometric identities $\sec^2 A = 1 + \tan^2 A$ and $\text{cosec}^2 A = 1 + \cot^2 A$.

Substitute these into the numerator and denominator:

$\frac{\sec^2 A}{\text{cosec}^2 A}$

Express $\sec^2 A$ and $\text{cosec}^2 A$ in terms of $\sin A$ and $\cos A$:

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

$\text{cosec}^2 A = \left(\frac{1}{\sin A}\right)^2 = \frac{1}{\sin^2 A}$

Substitute these back into the expression:

$\frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}}$

Invert the denominator and multiply:

$\frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1}$

$= \frac{\sin^2 A}{\cos^2 A}$

We know that $\frac{\sin A}{\cos A} = \tan A$. Therefore, $\frac{\sin^2 A}{\cos^2 A} = \left(\frac{\sin A}{\cos A}\right)^2 = \tan^2 A$.

The value of the expression is $\tan^2 A$.

Comparing with the options, the correct option is (D).

The correct option is (D).

Justification: By using the identities $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \text{cosec}^2 A$, and then converting to $\sin A$ and $\cos A$, the expression simplifies to $\tan^2 A$.

(i) Prove that $(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$:

Consider the Left Hand Side (LHS):

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

Express $\text{cosec } \theta$ and $\cot \theta$ in terms of $\sin \theta$ and $\cos \theta$:

$\text{cosec } \theta = \frac{1}{\sin \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$.

Substitute these into the LHS:

LHS = $\left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2$

Combine the terms inside the parenthesis:

LHS = $\left(\frac{1 - \cos \theta}{\sin \theta}\right)^2$

Square the numerator and the denominator separately:

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

Use the identity $\sin^2 \theta + \cos^2 \theta = 1$, which implies $\sin^2 \theta = 1 - \cos^2 \theta$.

Substitute $1 - \cos^2 \theta$ for $\sin^2 \theta$ in the denominator:

LHS = $\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$

Apply the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$, to the denominator, where $a=1$ and $b=\cos \theta$: $1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$.

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

Assume $1 - \cos \theta \neq 0$, which means $\cos \theta \neq 1$ (i.e., $\theta \neq 0^\circ, 360^\circ, ...$). For acute angles, this is true.

Cancel the common factor $(1 - \cos \theta)$ from the numerator and the denominator:

LHS = $\frac{1 - \cos \theta}{1 + \cos \theta}$

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

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(ii) Prove that $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$:

Consider the Left Hand Side (LHS):

LHS = $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}$

Find a common denominator, which is $(1 + \sin A)(\cos A)$:

LHS = $\frac{(\cos A)(\cos A) + (1 + \sin A)(1 + \sin A)}{(1 + \sin A)(\cos A)}$

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

Expand $(1 + \sin A)^2$ using $(a+b)^2 = a^2 + 2ab + b^2$:

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

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

Rearrange the terms in the numerator and use the identity $\sin^2 A + \cos^2 A = 1$:

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

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

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

Factor out 2 from the numerator:

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

Assume $1 + \sin A \neq 0$ and $\cos A \neq 0$. For acute angles, this is true.

Cancel the common factor $(1 + \sin A)$ from the numerator and the denominator:

LHS = $\frac{2}{\cos A}$

Use the reciprocal identity $\sec A = \frac{1}{\cos A}$:

LHS = $2 \times \frac{1}{\cos A} = 2 \sec A$

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

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(iii) Prove that $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{ cosec } \theta$:

Consider the Left Hand Side (LHS):

LHS = $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$

As hinted, write the expression in terms of $\sin \theta$ and $\cos \theta$.

$\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$.

Substitute these into the LHS:

LHS = $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$

Simplify the denominators by finding a common denominator:

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

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

Substitute these back into the LHS:

LHS = $\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$

Invert the denominators and multiply:

LHS = $\frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta - \cos \theta} + \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta - \sin \theta}$

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

Notice that $(\cos \theta - \sin \theta) = - (\sin \theta - \cos \theta)$. Factor out -1 from the denominator of the second term:

LHS = $\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta [- (\sin \theta - \cos \theta)]}$

LHS = $\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$

Now the denominators are $\cos \theta (\sin \theta - \cos \theta)$ and $\sin \theta (\sin \theta - \cos \theta)$. The common denominator is $\sin \theta \cos \theta (\sin \theta - \cos \theta)$.

LHS = $\frac{\sin^2 \theta \times \sin \theta - \cos^2 \theta \times \cos \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

LHS = $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

Apply the difference of cubes formula to the numerator: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, where $a = \sin \theta$ and $b = \cos \theta$.

Numerator = $(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)$

Use the identity $\sin^2 \theta + \cos^2 \theta = 1$ in the numerator:

Numerator = $(\sin \theta - \cos \theta)(1 + \sin \theta \cos \theta)$

Substitute the numerator back into the LHS expression:

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

Assume $\sin \theta - \cos \theta \neq 0$ (i.e., $\tan \theta \neq 1$, $\theta \neq 45^\circ$, etc.). For acute angles in general, this is true.

Cancel the common factor $(\sin \theta - \cos \theta)$ from the numerator and the denominator:

LHS = $\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$

Split the fraction into two terms:

LHS = $\frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$

Use the reciprocal identities $\frac{1}{\sin \theta} = \text{cosec } \theta$ and $\frac{1}{\cos \theta} = \sec \theta$. The second term simplifies to 1.

LHS = $\text{cosec } \theta \sec \theta + 1$

Rearrange the terms:

LHS = $1 + \sec \theta \text{ cosec } \theta$

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

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(iv) Prove that $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A}$:

As hinted, simplify LHS and RHS separately.

Consider the Left Hand Side (LHS):

LHS = $\frac{1 + \sec A}{\sec A}$

Split the fraction:

LHS = $\frac{1}{\sec A} + \frac{\sec A}{\sec A}$

Use the reciprocal identity $\cos A = \frac{1}{\sec A}$. The second term simplifies to 1.

LHS = $\cos A + 1$

Rearrange:

LHS = $1 + \cos A$

Now, consider the Right Hand Side (RHS):

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

Use the identity $\sin^2 A + \cos^2 A = 1$, which implies $\sin^2 A = 1 - \cos^2 A$.

Substitute $1 - \cos^2 A$ for $\sin^2 A$ in the numerator:

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

Apply the difference of squares formula to the numerator: $a^2 - b^2 = (a-b)(a+b)$, where $a=1$ and $b=\cos A$: $1 - \cos^2 A = (1 - \cos A)(1 + \cos A)$.

RHS = $\frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A}$

Assume $1 - \cos A \neq 0$, which means $\cos A \neq 1$ (i.e., $A \neq 0^\circ, 360^\circ, ...$). For acute angles, this is true unless A=0, but tan is undefined at 90, so we usually assume 0

Cancel the common factor $(1 - \cos A)$ from the numerator and the denominator:

RHS = $1 + \cos A$

Since LHS = $1 + \cos A$ and RHS = $1 + \cos A$, we have:

LHS = RHS

Hence Proved.

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(v) Prove that $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \text{cosec } A + \cot A$, using the identity $\text{cosec}^2 A = 1 + \cot^2 A$.

Consider the Left Hand Side (LHS):

LHS = $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1}$

To use the identity involving $\cot A$ and $\text{cosec } A$, divide the numerator and the denominator by $\sin A$, assuming $\sin A \neq 0$.

LHS = $\frac{\frac{\cos A}{\sin A} – \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} – \frac{1}{\sin A}}$

Using the identities $\cot A = \frac{\cos A}{\sin A}$ and $\text{cosec } A = \frac{1}{\sin A}$:

LHS = $\frac{\cot A – 1 + \text{cosec } A}{\cot A + 1 – \text{cosec } A}$

Rearrange the terms in the numerator:

LHS = $\frac{\cot A + \text{cosec } A – 1}{\cot A – \text{cosec } A + 1}$

We are asked to use the identity $\text{cosec}^2 A = 1 + \cot^2 A$. This identity can be rearranged to $1 = \text{cosec}^2 A - \cot^2 A$.

Using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$, we have $\text{cosec}^2 A - \cot^2 A = (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)$.

So, $1 = (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)$.

Substitute this expression for '1' into the numerator of the LHS:

Numerator = $(\cot A + \text{cosec } A) – 1$

Numerator = $(\cot A + \text{cosec } A) – (\text{cosec}^2 A - \cot^2 A)$

Numerator = $(\cot A + \text{cosec } A) – (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)$

Now, factor out the common term $(\cot A + \text{cosec } A)$ from the numerator:

Numerator = $(\cot A + \text{cosec } A) [1 – (\text{cosec } A - \cot A)]$

Numerator = $(\cot A + \text{cosec } A) [1 – \text{cosec } A + \cot A]$

Numerator = $(\cot A + \text{cosec } A) [\cot A – \text{cosec } A + 1]$

Substitute this back into the LHS expression:

LHS = $\frac{(\cot A + \text{cosec } A) (\cot A – \text{cosec } A + 1)}{\cot A – \text{cosec } A + 1}$

Assume $\cot A – \text{cosec } A + 1 \neq 0$. Cancel the common factor from the numerator and the denominator:

LHS = $\cot A + \text{cosec } A$

Rearrange the terms:

LHS = $\text{cosec } A + \cot A$

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

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(vi) Prove that $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$:

Consider the Left Hand Side (LHS):

LHS = $\sqrt{\frac{1 + \sin A}{1 - \sin A}}$

Rationalize the denominator inside the square root by multiplying the numerator and denominator by the conjugate of the denominator, which is $1 + \sin A$:

LHS = $\sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}}$

LHS = $\sqrt{\frac{(1 + \sin A)^2}{1^2 - \sin^2 A}}$

Use the identity $1 - \sin^2 A = \cos^2 A$ in the denominator:

LHS = $\sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$

Take the square root of the numerator and the denominator:

LHS = $\frac{\sqrt{(1 + \sin A)^2}}{\sqrt{\cos^2 A}}$

LHS = $\frac{|1 + \sin A|}{|\cos A|}$

For acute angle A ($0^\circ \leq A < 90^\circ$), $\sin A \geq 0$ and $\cos A \geq 0$. Thus $1 + \sin A$ is positive and $\cos A$ is positive.

LHS = $\frac{1 + \sin A}{\cos A}$

Split the fraction into two terms:

LHS = $\frac{1}{\cos A} + \frac{\sin A}{\cos A}$

Use the identities $\sec A = \frac{1}{\cos A}$ and $\tan A = \frac{\sin A}{\cos A}$:

LHS = $\sec A + \tan A$

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

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(vii) Prove that $\frac{\sin \theta – 2\sin^3 \theta}{2 \cos^3 \theta – \cos \theta} = \tan \theta$:

Consider the Left Hand Side (LHS):

LHS = $\frac{\sin \theta – 2\sin^3 \theta}{2 \cos^3 \theta – \cos \theta}$

Factor out $\sin \theta$ from the numerator and $\cos \theta$ from the denominator:

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

We know that $\frac{\sin \theta}{\cos \theta} = \tan \theta$. So, the expression is $\tan \theta \times \frac{1 – 2\sin^2 \theta}{2 \cos^2 \theta – 1}$.

Now we need to show that $\frac{1 – 2\sin^2 \theta}{2 \cos^2 \theta – 1} = 1$.

Use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to express the numerator or the denominator in terms of a single trigonometric function.

Let's express the numerator in terms of $\cos \theta$ by using $\sin^2 \theta = 1 - \cos^2 \theta$:

Numerator of the fraction = $1 – 2\sin^2 \theta = 1 – 2(1 - \cos^2 \theta)$

$= 1 – 2 + 2\cos^2 \theta = 2\cos^2 \theta - 1$

Substitute this back into the fraction part of the LHS:

Fraction part = $\frac{2\cos^2 \theta - 1}{2 \cos^2 \theta – 1}$

Assuming $2 \cos^2 \theta - 1 \neq 0$, we can cancel the terms:

Fraction part = 1

So, LHS = $\tan \theta \times 1$

LHS = $\tan \theta$

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

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(viii) Prove that $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$:

Consider the Left Hand Side (LHS):

LHS = $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2$}

Expand the squares using $(a+b)^2 = a^2 + 2ab + b^2$:

$(\sin A + \text{cosec } A)^2 = \sin^2 A + 2 \sin A \text{ cosec } A + \text{cosec}^2 A$

$(\cos A + \sec A)^2 = \cos^2 A + 2 \cos A \sec A + \sec^2 A$

Substitute these expanded forms back into the LHS:

LHS = $(\sin^2 A + 2 \sin A \text{ cosec } A + \text{cosec}^2 A) + (\cos^2 A + 2 \cos A \sec A + \sec^2 A)$

Rearrange the terms:

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 identities $\sin^2 A + \cos^2 A = 1$, $\sin A \text{ cosec } A = \sin A \times \frac{1}{\sin A} = 1$, and $\cos A \sec A = \cos A \times \frac{1}{\cos A} = 1$:

LHS = $1 + 2(1) + 2(1) + \text{cosec}^2 A + \sec^2 A$

LHS = $1 + 2 + 2 + \text{cosec}^2 A + \sec^2 A$

LHS = $5 + \text{cosec}^2 A + \sec^2 A$}

We need the RHS in terms of $\tan^2 A$ and $\cot^2 A$. Use the identities $\text{cosec}^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$:

LHS = $5 + (1 + \cot^2 A) + (1 + \tan^2 A)$

LHS = $5 + 1 + \cot^2 A + 1 + \tan^2 A$

LHS = $7 + \tan^2 A + \cot^2 A$

This is equal to the Right Hand Side (RHS).

LHS = RHS

Hence Proved.

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(ix) Prove that $(\text{cosec } A – \sin A)(\sec A – \cos A) = \frac{1}{\tan A + \cot A}$:

As hinted, simplify LHS and RHS separately.

Consider the Left Hand Side (LHS):

LHS = $(\text{cosec } A – \sin A)(\sec A – \cos A)$

Express $\text{cosec } A$ and $\sec A$ in terms of $\sin A$ and $\cos A$:

$\text{cosec } A = \frac{1}{\sin A}$ and $\sec A = \frac{1}{\cos A}$.

Substitute these into the LHS:

LHS = $\left(\frac{1}{\sin A} – \sin A\right)\left(\frac{1}{\cos A} – \cos A\right)$

Find a common denominator within each parenthesis:

$\frac{1}{\sin A} – \sin A = \frac{1}{\sin A} – \frac{\sin^2 A}{\sin A} = \frac{1 - \sin^2 A}{\sin A}$

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

Use the identities $1 - \sin^2 A = \cos^2 A$ and $1 - \cos^2 A = \sin^2 A$:

Substitute these back into the LHS expression:

LHS = $\left(\frac{\cos^2 A}{\sin A}\right)\left(\frac{\sin^2 A}{\cos A}\right)$

Multiply the two fractions:

LHS = $\frac{\cos^2 A \sin^2 A}{\sin A \cos A}$

Assume $\sin A \neq 0$ and $\cos A \neq 0$. Cancel common terms:

LHS = $\frac{(\cos A \cancel{\cos A}) (\sin A \cancel{\sin A})}{\cancel{\sin A} \cancel{\cos A}}$

LHS = $\cos A \sin A$

Now, consider the Right Hand Side (RHS):

RHS = $\frac{1}{\tan A + \cot A}$

Express $\tan A$ and $\cot A$ in terms of $\sin A$ and $\cos A$:

$\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$.

Substitute these into the denominator of the RHS:

RHS = $\frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$

Find a common denominator in the denominator of the main fraction. The common denominator is $\sin A \cos A$.

$\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin A \times \sin A}{\cos A \times \sin A} + \frac{\cos A \times \cos A}{\sin A \times \cos A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A}$

Use the identity $\sin^2 A + \cos^2 A = 1$ in the numerator:

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

Substitute this back into the RHS expression:

RHS = $\frac{1}{\frac{1}{\sin A \cos A}}$

Invert the denominator and multiply:

RHS = $1 \times (\sin A \cos A)$

RHS = $\sin A \cos A$

Since LHS = $\cos A \sin A$ and RHS = $\sin A \cos A$, we have:

LHS = RHS

Hence Proved.

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(x) Prove that $\left( \frac{1 + \tan^2 A}{1 + \cot^2 A} \right) = \left(\frac{1 - \tan A}{1 - \cot A} \right)^{2} = \tan^2 A$:

This identity has two parts to prove:

Part 1: $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A$

Part 2: $\left(\frac{1 - \tan A}{1 - \cot A} \right)^{2} = \tan^2 A$

If both parts are shown to be equal to $\tan^2 A$, then all three parts are equal to each other.

Part 1: Prove $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A$:

Consider the Left Hand Side (LHS):

LHS = $\frac{1 + \tan^2 A}{1 + \cot^2 A}$

Use the identities $\sec^2 A = 1 + \tan^2 A$ and $\text{cosec}^2 A = 1 + \cot^2 A$.

Substitute these into the LHS:

LHS = $\frac{\sec^2 A}{\text{cosec}^2 A}$

Express $\sec^2 A$ and $\text{cosec}^2 A$ in terms of $\sin A$ and $\cos A$:

$\sec^2 A = \frac{1}{\cos^2 A}$ and $\text{cosec}^2 A = \frac{1}{\sin^2 A}$.

Substitute these into the LHS:

LHS = $\frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}}$

Invert the denominator and multiply:

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

Using the identity $\frac{\sin A}{\cos A} = \tan A$, we have $\frac{\sin^2 A}{\cos^2 A} = \tan^2 A$.

LHS = $\tan^2 A$

This is equal to the Right Hand Side (RHS) of the first part.

Part 1 is Proved.

Part 2: Prove $\left(\frac{1 - \tan A}{1 - \cot A} \right)^{2} = \tan^2 A$:

Consider the Left Hand Side (LHS):

LHS = $\left(\frac{1 - \tan A}{1 - \cot A} \right)^{2}$

Express $\cot A$ in terms of $\tan A$ using the identity $\cot A = \frac{1}{\tan A}$.

Substitute this into the denominator of the fraction inside the square:

LHS = $\left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}} \right)^{2}$

Simplify the denominator of the inner fraction by finding a common denominator:

$1 - \frac{1}{\tan A} = \frac{\tan A - 1}{\tan A}$

Substitute this back into the LHS expression:

LHS = $\left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}} \right)^{2}$

Invert the denominator of the main fraction and multiply:

LHS = $\left((1 - \tan A) \times \frac{\tan A}{\tan A - 1} \right)^{2}$

Notice that $(1 - \tan A) = - (\tan A - 1)$. Substitute this into the expression:

LHS = $\left(- (\tan A - 1) \times \frac{\tan A}{\tan A - 1} \right)^{2}$}

Assume $\tan A - 1 \neq 0$ (i.e., $\tan A \neq 1$, $A \neq 45^\circ$, etc.). Cancel the common factor $(\tan A - 1)$:

LHS = $(- \tan A)^2$

LHS = $(-\tan A) \times (-\tan A) = \tan^2 A$

This is equal to the Right Hand Side (RHS) of the second part.

Part 2 is Proved.

Since both parts are equal to $\tan^2 A$, the given identity is true.

Hence Proved.