---
title: "NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1"
url: https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-1
dateModified: 2026-10-07T15:53:32+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1

Chapter 8: Introduction to Trigonometry. Every question from Exercise 8.1, 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 8.1

### 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.

## Related pages

- [All Chapter 8 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions)
- [Exercise 8.2](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-2)
- [Exercise 8.3](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions/exercise-8-3)

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