---
title: "NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2"
url: https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-2
dateModified: 2026-10-07T15:53:32+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2

Chapter 8: Introduction to Trigonometry. Every question from Exercise 8.2, with full working and the final answer.

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 8.2

### 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.

## Related pages

- [All Chapter 8 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions)
- [Exercise 8.1](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-1)
- [Exercise 8.3](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-3)

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