---
title: "NCERT Solutions Class 10 Maths Ch 8 Introduction to Trigonometry"
url: https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions
dateModified: 2026-10-07T15:53:32+00:00
---

# NCERT Solutions Class 10 Maths Ch 8 Introduction to Trigonometry

This chapter's questions cover the fundamentals of trigonometric ratios, their values for specific angles, and the application of trigonometric identities for acute angles in right-angled triangles.

Free PDF (14 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-b7f11c4718.pdf

## EXERCISE 8.1

### Question 1

*3 marks · Short answer*

In $\Delta ABC$, right-angled at $B$, $AB = 24\text{ cm}$, $BC = 7\text{ cm}$. Determine :
(i) $\sin A$, $\cos A$
(ii) $\sin C$, $\cos C$

**Part (i)**

1. In $\Delta ABC$, right-angled at $B$, $AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625$.
2. Therefore, $AC = \sqrt{625} = 25\text{ cm}$.
3. $\sin A = \frac{\text{BC}}{\text{AC}} = \frac{7}{25}$ and $\cos A = \frac{\text{AB}}{\text{AC}} = \frac{24}{25}$.

Answer (i): $\sin A = \frac{7}{25}$, $\cos A = \frac{24}{25}$

**Part (ii)**

1. $\sin C = \frac{\text{AB}}{\text{AC}} = \frac{24}{25}$ and $\cos C = \frac{\text{BC}}{\text{AC}} = \frac{7}{25}$.

Answer (ii): $\sin C = \frac{24}{25}$, $\cos C = \frac{7}{25}$

**Answer:** (i) $\sin A = \frac{7}{25}$, $\cos A = \frac{24}{25}$ (ii) $\sin C = \frac{24}{25}$, $\cos C = \frac{7}{25}$

> Common mistake: Confusing the sides opposite and adjacent when the angle changes from A to C.

### Question 2

*3 marks · Short answer*

In Fig. 8.13, find $\tan P - \cot R$.

**Solution**

1. In right $\Delta PQR$, by Pythagoras theorem, $PR^2 = PQ^2 + QR^2$.
2. $13^2 = 12^2 + QR^2 \implies 169 = 144 + QR^2 \implies QR^2 = 25 \implies QR = 5\text{ cm}$.
3. $\tan P = \frac{QR}{PQ} = \frac{5}{12}$ and $\cot R = \frac{QR}{PQ} = \frac{5}{12}$.
4. $\tan P - \cot R = \frac{5}{12} - \frac{5}{12} = 0$.

**Answer:** $0$

> Common mistake: Using $\cot R$ incorrectly as $\frac{PQ}{QR}$ instead of $\frac{\text{side adjacent to } R}{\text{side opposite to } R}$.

### Question 3

*3 marks · Short answer*

If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.

**Solution**

1. Given $\sin A = \frac{3}{4} = \frac{\text{BC}}{\text{AC}}$, let $BC = 3k$ and $AC = 4k$, where $k$ is a positive number.
2. By Pythagoras theorem, $AB^2 = AC^2 - BC^2 = (4k)^2 - (3k)^2 = 16k^2 - 9k^2 = 7k^2$.
3. Therefore, $AB = \sqrt{7}k$.
4. $\cos A = \frac{\text{AB}}{\text{AC}} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4}$ and $\tan A = \frac{\text{BC}}{\text{AB}} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}}$.

**Answer:** $\cos A = \frac{\sqrt{7}}{4}$, $\tan A = \frac{3}{\sqrt{7}}$

> Common mistake: Forgetting to introduce the positive constant $k$ when working with ratios of sides.

### Question 4

*3 marks · Short answer*

Given $15 \cot A = 8$, find $\sin A$ and $\sec A$.

**Solution**

1. Given $15 \cot A = 8$, so $\cot A = \frac{8}{15} = \frac{\text{AB}}{\text{BC}}$.
2. Let $AB = 8k$ and $BC = 15k$, where $k$ is a positive number.
3. By Pythagoras theorem, $AC^2 = AB^2 + BC^2 = (8k)^2 + (15k)^2 = 64k^2 + 225k^2 = 289k^2$.
4. Therefore, $AC = \sqrt{289}k = 17k$.
5. $\sin A = \frac{\text{BC}}{\text{AC}} = \frac{15k}{17k} = \frac{15}{17}$ and $\sec A = \frac{\text{AC}}{\text{AB}} = \frac{17k}{8k} = \frac{17}{8}$.

**Answer:** $\sin A = \frac{15}{17}$, $\sec A = \frac{17}{8}$

> Common mistake: Wrongly equating $\cot A$ to opposite over adjacent instead of adjacent over opposite.

### Question 5

*3 marks · Short answer*

Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.

**Solution**

1. Given $\sec \theta = \frac{13}{12} = \frac{\text{Hypotenuse}}{\text{Base}}$. Let Hypotenuse $= 13k$ and Base $= 12k$.
2. By Pythagoras theorem, $\text{Perpendicular} = \sqrt{(13k)^2 - (12k)^2} = \sqrt{25k^2} = 5k$.
3. $\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5k}{13k} = \frac{5}{13}$
4. $\cos \theta = \frac{1}{\sec \theta} = \frac{12}{13}$, $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{5}{12}$
5. $\text{cosec } \theta = \frac{1}{\sin \theta} = \frac{13}{5}$ and $\cot \theta = \frac{1}{\tan \theta} = \frac{12}{5}$.

**Answer:** $\sin \theta = \frac{5}{13}$, $\cos \theta = \frac{12}{13}$, $\tan \theta = \frac{5}{12}$, $\text{cosec } \theta = \frac{13}{5}$, $\cot \theta = \frac{12}{5}$

> Common mistake: Mixing up the definitions of secant and cosine ratios.

### Question 6

*3 marks · Proof*

If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.

**Solution**

1. Let us consider two right triangles $ABC$ and $PQR$ where $\angle C = 90^\circ$ and $\angle R = 90^\circ$.
2. Given that $\cos A = \cos P$.
3. From the triangles, $\cos A = \frac{AC}{AB}$ and $\cos P = \frac{PR}{PQ}$.
4. Therefore, $\frac{AC}{AB} = \frac{PR}{PQ}$, which can be written as $\frac{AC}{PR} = \frac{AB}{PQ} = k$ (say).
5. By Pythagoras theorem, $BC = \sqrt{AB^2 - AC^2}$ and $QR = \sqrt{PQ^2 - PR^2} = \sqrt{k^2 AB^2 - k^2 AC^2} = k \sqrt{AB^2 - AC^2} = k BC$.
6. Thus, $\frac{BC}{QR} = \frac{AB}{PQ} = \frac{AC}{PR} = k$.
7. By SSS similarity criterion, $\Delta ABC \sim \Delta PQR$.
8. Therefore, $\angle A = \angle P$ (corresponding parts of similar triangles are equal).

**Answer:** Hence proved.

> Common mistake: Assuming triangles are congruent instead of similar when dealing with ratios of sides.

### Question 7

*3 marks · Case-based*

If $\cot \theta = \frac{7}{8}$, evaluate :
(i) $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$
(ii) $\cot^2 \theta$

**Part (i)**

1. Given $\cot \theta = \frac{7}{8}$
2. The expression is $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)} = \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = (\cot \theta)^2$
3. Substitute $\cot \theta = \frac{7}{8}$ into the simplified expression to get $\left(\frac{7}{8}\right)^2$
4. Result: $\frac{49}{64}$

Answer (i): $\frac{49}{64}$

**Part (ii)**

1. Given $\cot \theta = \frac{7}{8}$
2. We need to evaluate $\cot^2 \theta$
3. Substitute $\cot \theta = \frac{7}{8}$ to get $\left(\frac{7}{8}\right)^2$
4. Result: $\frac{49}{64}$

Answer (ii): $\frac{49}{64}$

**Answer:** (i) $\frac{49}{64}$, (ii) $\frac{49}{64}$

> Common mistake: Expanding the terms using a right triangle without noticing the algebraic identity $(1+\sin\theta)(1-\sin\theta) = 1-\sin^2\theta = \cos^2\theta$.

### Question 8

*3 marks · Proof*

If $3 \cot A = 4$, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ or not.

**Solution**

1. Given $3 \cot A = 4$, so $\cot A = \frac{4}{3}$
2. Let a right-angled triangle $ABC$ have $\tan A = \frac{1}{\cot A} = \frac{3}{4} = \frac{BC}{AB}$
3. Let $BC = 3k$ and $AB = 4k$, where $k$ is a positive number
4. By Pythagoras theorem, $AC = \sqrt{AB^2 + BC^2} = \sqrt{(4k)^2 + (3k)^2} = 5k$
5. Therefore, $\sin A = \frac{BC}{AC} = \frac{3}{5}$ and $\cos A = \frac{AB}{AC} = \frac{4}{5}$
6. LHS = $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$
7. RHS = $\cos^2 A - \sin^2 A = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$
8. Since LHS = RHS, the given statement is true.

**Answer:** Yes, $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$

> Common mistake: Forgetting to introduce the positive constant $k$ when taking the ratio of sides.

### Question 9

*3 marks · Case-based*

In triangle $ABC$, right-angled at $B$, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of:
(i) $\sin A \cos C + \cos A \sin C$
(ii) $\cos A \cos C - \sin A \sin C$

**Part (i)**

1. Given $\tan A = \frac{1}{\sqrt{3}} = \frac{BC}{AB}$
2. Let $BC = 1k$ and $AB = \sqrt{3}k$, where $k$ is a positive number
3. By Pythagoras theorem, $AC = \sqrt{AB^2 + BC^2} = \sqrt{(\sqrt{3}k)^2 + (1k)^2} = \sqrt{4k^2} = 2k$
4. Therefore, $\sin A = \frac{1}{2}$, $\cos A = \frac{\sqrt{3}}{2}$, $\sin C = \frac{\sqrt{3}}{2}$, and $\cos C = \frac{1}{2}$
5. Substitute the values into $\sin A \cos C + \cos A \sin C = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(/\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{4} + \frac{3}{4} = 1$

Answer (i): $1$

**Part (ii)**

1. From the triangle sides, $\cos A = \frac{\sqrt{3}}{2}$, $\cos C = \frac{1}{2}$, $\sin A = \frac{1}{2}$, and $\sin C = \frac{\sqrt{3}}{2}$
2. Substitute the values into $\cos A \cos C - \sin A \sin C = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$
3. Result: $\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0$

Answer (ii): $0$

**Answer:** (i) $1$, (ii) $0$

> Common mistake: Confusing the sides opposite to angle A and angle C.

### Question 10

*3 marks · Short answer*

In $\Delta PQR$, right-angled at $Q$, $PR + QR = 25\text{ cm}$ and $PQ = 5\text{ cm}$. Determine the values of $\sin P$, $\cos P$ and $\tan P$.

**Solution**

1. Given $\Delta PQR$ is right-angled at $Q$, $PQ = 5\text{ cm}$, and $PR + QR = 25\text{ cm}$
2. Let $QR = x$, then $PR = 25 - x$
3. By Pythagoras theorem, $PR^2 = PQ^2 + QR^2$
4. Substitute the values: $(25 - x)^2 = 5^2 + x^2$
5. Expand and simplify: $625 - 50x + x^2 = 25 + x^2 \implies 50x = 600 \implies x = 12\text{ cm}$
6. Thus, $QR = 12\text{ cm}$ and $PR = 25 - 12 = 13\text{ cm}$
7. Determine the ratios: $\sin P = \frac{QR}{PR} = \frac{12}{13}$, $\cos P = \frac{PQ}{PR} = \frac{5}{13}$, and $\tan P = \frac{QR}{PQ} = \frac{12}{5}$

**Answer:** $\sin P = \frac{12}{13}$, $\cos P = \frac{5}{13}$, $\tan P = \frac{12}{5}$

> Common mistake: Wrong expansion of $(25-x)^2$ leading to incorrect side lengths.

### Question 11

*1 mark · True or false*

State whether the following are true or false. Justify your answer.
(i) The value of $\tan A$ is always less than $1$.
(ii) $\sec A = \frac{12}{5}$ for some value of angle $A$.
(iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.
(iv) $\cot A$ is the product of $\cot$ and $A$.
(v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.

**Part (i)**

1. Statement: The value of $\tan A$ is always less than $1$.
2. Reason: $\tan A$ can take any real value, and for angles greater than $45^\circ$, its value is greater than $1$.

Answer (i): False

**Part (ii)**

1. Statement: $\sec A = \frac{12}{5}$ for some value of angle $A$.
2. Reason: Since $\sec A = \frac{\text{hypotenuse}}{\text{base}}$ and the hypotenuse is the longest side, $\sec A$ can be greater than $1$.

Answer (ii): True

**Part (iii)**

1. Statement: $\cos A$ is the abbreviation used for the cosecant of angle $A$.
2. Reason: $\cos A$ is the abbreviation for cosine of angle $A$, whereas cosecants are abbreviated as $\operatorname{cosec} A$.

Answer (iii): False

**Part (iv)**

1. Statement: $\cot A$ is the product of $\cot$ and $A$.
2. Reason: $\cot A$ is a single notation representing the cotangent of angle $A$ and is not a product.

Answer (iv): False

**Part (v)**

1. Statement: $\sin \theta = \frac{4}{3}$ for some angle $\theta$.
2. Reason: Since $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ and hypotenuse is always greater than or equal to the opposite side, $\sin \theta$ can never exceed $1$.

Answer (v): False

**Answer:** (i) False, (ii) True, (iii) False, (iv) False, (v) False

> Common mistake: Confusing secant and cosine abbreviations, or forgetting that sine values cannot exceed 1.

## EXERCISE 8.2

### Question 1

*3 marks · Short answer*

Evaluate the following :
(i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
(ii) $2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$
(iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}$
(iv) $\frac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
(v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$

**Part (i)**

1. Substitute standard values: $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 30^\circ = \frac{1}{2}$, $\cos 60^\circ = \frac{1}{2}$.
2. Calculate: $\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4}$.
3. Simplify: $\frac{4}{4} = 1$.

Answer (i): 1

**Part (ii)**

1. Substitute standard values: $\tan 45^\circ = 1$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$.
2. Calculate: $2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2$.
3. Simplify: $2 + \frac{3}{4} - \frac{3}{4} = 2$.

Answer (ii): 2

**Part (iii)**

1. Substitute standard values: $\cos 45^\circ = \frac{1}{\sqrt{2}}$, $\sec 30^\circ = \frac{2}{\sqrt{3}}$, $\csc 30^\circ = 2$.
2. Simplify denominator: $\frac{2}{\sqrt{3}} + 2 = \frac{2 + 2\sqrt{3}}{\sqrt{3}}$.
3. Divide and rationalize: $\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1 + \sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3} + 1)} = \frac{3\sqrt{2} - \sqrt{6}}{8}$.

Answer (iii): \frac{3\sqrt{2} - \sqrt{6}}{8}

**Part (iv)**

1. Substitute standard values: $\sin 30^\circ = \frac{1}{2}$, $\tan 45^\circ = 1$, $\csc 60^\circ = \frac{2}{\sqrt{3}}$, $\sec 30^\circ = \frac{2}{\sqrt{3}}$, $\cos 60^\circ = \frac{1}{2}$, $\cot 45^\circ = 1$.
2. Numerator: $\frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}} = \frac{3\sqrt{3} - 4}{2\sqrt{3}}$.
3. Denominator: $\frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{3}{2} + \frac{2}{\sqrt{3}} = \frac{3\sqrt{3} + 4}{2\sqrt{3}}$.
4. Divide numerator by denominator: $\frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$.

Answer (iv): \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}

**Part (v)**

1. Substitute standard values: $\cos 60^\circ = \frac{1}{2}$, $\sec 30^\circ = \frac{2}{\sqrt{3}}$, $\tan 45^\circ = 1$, $\sin 30^\circ = \frac{1}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
2. Numerator: $5\left(\frac{1}{4}\right) + 4\left(\frac{4}{3}\right) - 1 = \frac{5}{4} + \frac{16}{3} - 1 = \frac{15 + 64 - 12}{12} = \frac{67}{12}$.
3. Denominator: $\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$.
4. Calculate final result: $\frac{67}{12} \div 1 = \frac{67}{12}$.

Answer (v): \frac{67}{12}

**Answer:** Solutions evaluated for parts (i) to (v).

> Common mistake: Errors in simplifying complex fractions and rationalizing denominators.

### Question 2

*1 mark · MCQ*

Choose the correct option and justify your choice :
(i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} =$
(ii) $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} =$
(iii) $\sin 2A = 2 \sin A$ is true when $A =$
(iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} =$

- \sin 60^\circ, \cos 60^\circ, \tan 60^\circ, \sin 30^\circ
- \tan 90^\circ, 1, \sin 45^\circ, 0
- 0^\circ, 30^\circ, 45^\circ, 60^\circ
- \cos 60^\circ, \sin 60^\circ, \tan 60^\circ, \sin 30^\circ

**Part (i)**

1. Substitute $\tan 30^\circ = \frac{1}{\sqrt{3}}$ into the expression: $\frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3}$.
2. Simplify: $\frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2}$, which is equal to $\sin 60^\circ$.

Answer (i): (A) \sin 60^\circ

**Part (ii)**

1. Substitute $\tan 45^\circ = 1$ into the expression: $\frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1}$.
2. Simplify: $\frac{0}{2} = 0$.

Answer (ii): (D) 0

**Part (iii)**

1. Test $A = 0^\circ$: $\sin(2 \times 0^\circ) = \sin 0^\circ = 0$ and $2 \sin 0^\circ = 2(0) = 0$.
2. Since both sides are equal, the statement is true for $A = 0^\circ$.

Answer (iii): (A) 0^\circ

**Part (iv)**

1. Substitute $\tan 30^\circ = \frac{1}{\sqrt{3}}$ into the expression: $\frac{2(1/\sqrt{3})}{1 - (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 - 1/3} = \frac{2/\sqrt{3}}{2/3}$.
2. Simplify: $\frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3}$, which is equal to $\tan 60^\circ$.

Answer (iv): (C) \tan 60^\circ

**Answer:** Selected options for parts (i) to (iv).

> Common mistake: Confusing $\tan 60^\circ$ with $\sin 60^\circ$ or mixing up the plus and minus signs in parts (i) and (iv).

### Question 3

*3 marks · Short answer*

If $\tan(A + B) = \sqrt{3}$ and $\tan(A - B) = \frac{1}{\sqrt{3}}$; $0^\circ < A + B \le 90^\circ$; $A > B$, find $A$ and $B$.

**Solution**

1. We are given $\tan(A + B) = \sqrt{3}$.
2. Since $\tan 60^\circ = \sqrt{3}$, we get $A + B = 60^\circ$ (1)
3. We are also given $\tan(A - B) = \frac{1}{\sqrt{3}}$.
4. Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we get $A - B = 30^\circ$ (2)
5. Adding equations (1) and (2), we get $2A = 90^\circ$, which gives $A = 45^\circ$.
6. Substituting $A = 45^\circ$ in equation (1), we get $45^\circ + B = 60^\circ$, which gives $B = 15^\circ$.

**Answer:** $A = 45^\circ$ and $B = 15^\circ$

> Common mistake: Confusing the standard angle values for tangent, such as taking $\tan 30^\circ = \sqrt{3}$ instead of $\frac{1}{\sqrt{3}}$.

### Question 4

*1 mark · True or false*

State whether the following are true or false. Justify your answer.
(i) $\sin(A + B) = \sin A + \sin B$.
(ii) The value of $\sin \theta$ increases as $\theta$ increases.
(iii) The value of $\cos \theta$ increases as $\theta$ increases.
(iv) $\sin \theta = \cos \theta$ for all values of $\theta$.
(v) $\cot A$ is not defined for $A = 0^\circ$.

**Part (i)**

1. Consider $A = 30^\circ$ and $B = 60^\circ$. $\sin(30^\circ + 60^\circ) = \sin 90^\circ = 1$, whereas $\sin 30^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}$.
2. Since $1 \neq \frac{1 + \sqrt{3}}{2}$, the statement is false.

Answer (i): False

**Part (ii)**

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

Answer (ii): True

**Part (iii)**

1. As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\cos \theta$ decreases from $1$ to $0$, not increases.

Answer (iii): False

**Part (iv)**

1. This is true only for $\theta = 45^\circ$ where $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$. For other angles like $30^\circ$, $\sin 30^\circ \neq \cos 30^\circ$.

Answer (iv): False

**Part (v)**

1. By definition, $\cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0}$, which is not defined.

Answer (v): True

**Answer:** True/False evaluated for parts (i) to (v).

> Common mistake: Assuming trigonometric functions are distributive over addition.

## EXERCISE 8.3

### Question 1

*3 marks · Proof*

Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.

**Solution**

1. We know that $\tan A = \frac{1}{\cot A}$.
2. Also, $\sin A = \frac{1}{\csc A}$, and since $\csc^2 A = 1 + \cot^2 A$, we have $\csc A = \sqrt{1 + \cot^2 A}$.
3. Therefore, $\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$.
4. For $\sec A$, we use $\sec^2 A = 1 + \tan^2 A = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A}$.
5. Taking the square root, we get $\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot^2 A}$ (or $\frac{\sqrt{1 + \cot^2 A}}{\cot A}$ since angles are acute).
6. Hence expressed.

**Answer:** $\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$, $\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$, $\tan A = \frac{1}{\cot A}$

> Common mistake: Forgetting to take the square root properly when converting secant and sine.

### Question 2

*4 marks · Proof*

Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.

**Solution**

1. We know that $\cos A = \frac{1}{\sec A}$.
2. Using $\sin^2 A + \cos^2 A = 1$, we get $\sin^2 A = 1 - \cos^2 A = 1 - \frac{1}{\sec^2 A} = \frac{\sec^2 A - 1}{\sec^2 A}$, so $\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$.
3. For $\csc A$, taking the reciprocal of $\sin A$ gives $\csc A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$.
4. For $\tan A$, using $\tan^2 A = \sec^2 A - 1$, we get $\tan A = \sqrt{\sec^2 A - 1}$.
5. For $\cot A$, taking the reciprocal of $\tan A$ gives $\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$.
6. Hence all ratios are expressed in terms of $\sec A$.

**Answer:** $\cos A = \frac{1}{\sec A}$, $\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$, $\csc A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$, $\tan A = \sqrt{\sec^2 A - 1}$, $\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$

> Common mistake: Mixing up the reciprocals and square roots.

### Question 3

*1 mark · MCQ*

Choose the correct option. Justify your choice.
(i) $9 \sec^2 A - 9 \tan^2 A =$
(ii) $(1 + \tan \theta + \sec \theta)(1 + \cot \theta - \csc \theta) =$
(iii) $(\sec A + \tan A)(1 - \sin A) =$
(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$

- 1, 9, 8, 0
- 0, 1, 2, -1
- \sec A, \sin A, \csc A, \cos A
- \sec^2 A, -1, \cot^2 A, \tan^2 A

**Part (i)**

1. $9 \sec^2 A - 9 \tan^2 A = 9(\sec^2 A - \tan^2 A)$
2. Using the identity $\sec^2 A - \tan^2 A = 1$, we get $9(1) = 9$.

Answer (i): (B) 9

**Part (ii)**

1. Expressing in terms of sine and cosine: $\left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right)\left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)$
2. Simplifying each bracket: $\left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right)\left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right)$
3. Using $(a+b)(a-b)$ in the numerator: $\frac{(\cos \theta + \sin \theta)^2 - 1^2}{\sin \theta \cos \theta}$
4. Expanding and simplifying: $\frac{\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta} = \frac{1 + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta} = 2$

Answer (ii): (C) 2

**Part (iii)**

1. Converting to sine and cosine: $\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 - \sin A)$
2. Combining terms: $\left(\frac{1 + \sin A}{\cos A}\right)(1 - \sin A) = \frac{1 - \sin^2 A}{\cos A}$
3. Since $1 - \sin^2 A = \cos^2 A$, we get $\frac{\cos^2 A}{\cos A} = \cos A$.

Answer (iii): (D) \cos A

**Part (iv)**

1. Writing in terms of sine and cosine: $\frac{1 + \frac{\sin^2 A}{\cos^2 A}}{1 + \frac{\cos^2 A}{\sin^2 A}}$
2. Simplifying the numerator and denominator: $\frac{\frac{\cos^2 A + \sin^2 A}{\cos^2 A}}{\frac{\sin^2 A + \cos^2 A}{\sin^2 A}} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}}$
3. Reciprocating and multiplying: $\frac{\sin^2 A}{\cos^2 A} = \tan^2 A$

Answer (iv): (D) \tan^2 A

**Answer:** (i) (B), (ii) (B), (iii) (D), (iv) (D)

> Common mistake: Forgetting standard trigonometric identities like $\sec^2 A - \tan^2 A = 1$ or making algebraic expansion errors in part (ii).

### Question 4

*5 marks · Proof*

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) $(\csc \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
(ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$
(iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$
(iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A}$
(v) $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$
(vi) $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$
(vii) $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$
(viii) $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
(ix) $(\csc A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$
(x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

**Part (i)**

1. LHS $= (\csc \theta - \cot \theta)^2 = \left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$.
2. Substitute $\sin^2 \theta = 1 - \cos^2 \theta$: $\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)}$.
3. Cancel $(1 - \cos \theta)$ to get $\frac{1 - \cos \theta}{1 + \cos \theta} =$ RHS.
4. Hence proved.

Answer (i): Hence proved.

**Part (ii)**

1. LHS $= \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A}$.
2. Expand numerator: $\frac{\cos^2 A + 1 + \sin^2 A + 2 \sin A}{(1 + \sin A)\cos A} = \frac{(\cos^2 A + \sin^2 A) + 1 + 2 \sin A}{(1 + \sin A)\cos A}$.
3. Since $\cos^2 A + \sin^2 A = 1$, we get $\frac{1 + 1 + 2 \sin A}{(1 + \sin A)\cos A} = \frac{2 + 2 \sin A}{(1 + \sin A)\cos A} = \frac{2(1 + \sin A)}{(1 + \sin A)\cos A}$.
4. Cancel $(1 + \sin A)$ to get $\frac{2}{\cos A} = 2 \sec A =$ RHS.
5. Hence proved.

Answer (ii): Hence proved.

**Part (iii)**

1. Write in terms of sine and cosine: $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$.
2. Simplify fractions: $\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)} = \frac{1}{\sin \theta - \cos \theta} \left(\frac{\sin^3 \theta - \cos^3 \theta}{\cos \theta \sin \theta}\right)$.
3. Use algebraic formula for difference of cubes: $\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\cos \theta \sin \theta (\sin \theta - \cos \theta)} = \frac{1 + \sin \theta \cos \theta}{\cos \theta \sin \theta}$.
4. Split terms: $\frac{1}{\cos \theta \sin \theta} + \frac{\sin \theta \cos \theta}{\cos \theta \sin \theta} = \sec \theta \csc \theta + 1 =$ RHS.
5. Hence proved.

Answer (iii): Hence proved.

**Part (iv)**

1. LHS $= \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = 1 + \cos A$.
2. Multiply numerator and denominator by $(1 - \cos A)$: $\frac{(1 + \cos A)(1 - \cos A)}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{\sin^2 A}{1 - \cos A} =$ RHS.
3. Hence proved.

Answer (iv): Hence proved.

**Part (v)**

1. Divide numerator and denominator of LHS by $\sin A$: $\frac{\frac{\cos A}{\sin A} - 1 + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + 1 - \frac{1}{\sin A}} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A}$.
2. Rearrange terms using $\csc^2 A - \cot^2 A = 1$: $\frac{(\csc A + \cot A) - (\csc^2 A - \cot^2 A)}{\cot A + 1 - \csc A} = \frac{(\csc A + \cot A)[1 - (\csc A - \cot A)]}{\cot A + 1 - \csc A}$.
3. Simplify and cancel common bracket: $\frac{(\csc A + \cot A)(1 - \csc A + \cot A)}{1 - \csc A + \cot A} = \csc A + \cot A =$ RHS.
4. Hence proved.

Answer (v): Hence proved.

**Part (vi)**

1. Rationalize inside the square root: $\sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$.
2. Substitute $1 - \sin^2 A = \cos^2 A$: $\sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A}$.
3. Split terms: $\frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A =$ RHS.
4. Hence proved.

Answer (vi): Hence proved.

**Part (vii)**

1. Take out common factors: LHS $= \frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(2 \cos^2 \theta - 1)}$.
2. Substitute $\cos^2 \theta = 1 - \sin^2 \theta$ in the denominator: $\frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(2(1 - \sin^2 \theta) - 1)} = \frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(2 - 2 \sin^2 \theta - 1)}$.
3. Simplify denominator: $\frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(1 - 2 \sin^2 \theta)} = \frac{\sin \theta}{\cos \theta} = \tan \theta =$ RHS.
4. Hence proved.

Answer (vii): Hence proved.

**Part (viii)**

1. Expand both squares: LHS $= (\sin^2 A + \csc^2 A + 2 \sin A \csc A) + (\cos^2 A + \sec^2 A + 2 \cos A \sec A)$.
2. Group sine and cosine terms: $(\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 2(1) + 2(1)$.
3. Substitute identities $\sin^2 A + \cos^2 A = 1$, $\csc^2 A = 1 + \cot^2 A$, and $\sec^2 A = 1 + \tan^2 A$: $1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 4$.
4. Combine constants: $1 + 1 + 1 + 4 + \tan^2 A + \cot^2 A = 7 + \tan^2 A + \cot^2 A =$ RHS.
5. Hence proved.

Answer (viii): Hence proved.

**Part (ix)**

1. Simplify LHS: $\left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) = \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right) = \frac{\cos^2 A \sin^2 A}{\sin A \cos A} = \sin A \cos A$.
2. Simplify RHS: $\frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\cos A \sin A}} = \frac{1}{\frac{1}{\cos A \sin A}} = \cos A \sin A$.
3. Since LHS = RHS.
4. Hence proved.

Answer (ix): Hence proved.

**Part (x)**

1. Simplify the first term: $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$.
2. Simplify the second term: $\left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}}\right)^2 = \left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}\right)^2 = (-\tan A)^2 = \tan^2 A$.
3. Since all parts equal $\tan^2 A$.
4. Hence proved.

Answer (x): Hence proved.

**Answer:** All identities proved successfully.

> Common mistake: Algebraic expansion errors or incorrect application of square identities.

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry?

This chapter has three exercises with a total of 19 questions based on the 2026-27 NCERT textbook. Exercise 8.1 has 11 questions, Exercise 8.2 has 4 questions, and Exercise 8.3 has 4 questions. You can find SwaVid's free PDF and step-by-step solutions for all these questions on this page only.

### What topics do the questions in Exercise 8.1 cover?

Exercise 8.1 covers 11 questions featuring case, proof, short answer, and true or false types. The concepts include trigonometric ratios of a right triangle, angles opposite to equal trigonometric ratios, and calculating ratios using the Pythagoras theorem. SwaVid provides detailed step-by-step solutions for all these concepts on this page only.

### What is the hardest question type in this chapter and how should we approach it?

Proof-based questions, especially in Exercise 8.3, are often considered the most challenging by students. To approach them, you should carefully apply standard trigonometric identities and express terms in sine and cosine when stuck. SwaVid's free PDF on this page offers clear explanations to master these proofs.

### How can I write answers in Class 10 board exams to score full marks in trigonometry?

To secure full marks, always state the given formulas clearly and show every intermediate step of calculation without skipping. Drawing a neat right-angled triangle and labeling the sides properly for ratios like $\sin A$ or $\sec A$ is also essential. You can refer to SwaVid's step-by-step solutions available on this page to learn the correct presentation format.

### Is the free PDF for these Class 10 Maths Chapter 8 solutions available for download?

Yes, the complete free PDF containing detailed answers for all exercises is readily accessible to students. It covers specific angle values, standard identities, and trigonometric ratios as per the latest syllabus. You can download SwaVid's free PDF and view step-by-step solutions directly on this page.

## Related pages

- [Exercise 8.1 solutions](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-1)
- [Exercise 8.2 solutions](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-2)
- [Exercise 8.3 solutions](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-3)
- [Introduction to Trigonometry: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/introduction-to-trigonometry)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student