---
title: "Introduction to Trigonometry: CBSE Class 10 Maths previous year questions"
url: https://www.swavid.com/cbse/class-10/maths/pyq/introduction-to-trigonometry
---

# Introduction to Trigonometry: CBSE Class 10 Maths previous year questions

35 questions from CBSE Class 10 board papers, newest first, each with full working.

## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions

### Question 9

*1 mark · MCQ*

Given that $\sin \theta = \frac{a}{b}$, then $\cos \theta$ is equal to :

- \frac{b}{\sqrt{b^2 - a^2}}
- \frac{b}{a}
- \frac{\sqrt{b^2 - a^2}}{b}
- \frac{a}{\sqrt{b^2 - a^2}}

**Solution**

1. Given $\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{a}{b}$.
2. By using identity $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b}$.

**Answer:** (c) \frac{\sqrt{b^2 - a^2}}{b}

> Common mistake: Inverting the numerator and denominator.

### Question 10

*1 mark · MCQ*

If $\cos A = \frac{1}{2}$, then the value of $\sin^2 A + 2\cos^2 A$ is :

- \frac{3}{2}
- \frac{5}{4}
- -1
- \frac{1}{2}

**Solution**

1. Given $\cos A = \frac{1}{2}$, which implies $A = 60^\circ$.
2. Therefore, $\sin A = \frac{\sqrt{3}}{2}$, so $\sin^2 A = \frac{3}{4}$ and $\cos^2 A = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
3. Substitute these values into the expression: $\sin^2 A + 2\cos^2 A = \frac{3}{4} + 2\left(\frac{1}{4}\right) = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$.

**Answer:** (b) \frac{5}{4}

> Common mistake: Forgetting to square the trigonometric ratios or substituting incorrect values.

### Question 24

*2 marks · Very short answer*

(A) If $\tan \theta = \frac{24}{7}$, then find the value of $\sin \theta + \cos \theta$.

**Solution**

1. Given $\tan \theta = \frac{24}{7} = \frac{\text{Perpendicular}}{\text{Base}}$.
2. Let perpendicular be $24k$ and base be $7k$. By Pythagoras theorem, hypotenuse $=\sqrt{(24k)^2 + (7k)^2} = 25k$.
3. Therefore, $\sin \theta = \frac{24}{25}$ and $\cos \theta = \frac{7}{25}$.
4. Add the values: $\sin \theta + \cos \theta = \frac{24}{25} + \frac{7}{25} = \frac{31}{25}$.

**Answer:** $\frac{31}{25}$

> Common mistake: Mixing up the perpendicular and base sides for the given trigonometric ratio.

### Question 24 (OR)

*2 marks · Very short answer*

(B) If $\cot \theta = \frac{7}{8}$, then find the value of $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$.

**Solution**

1. Given expression is $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)} = \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta}$.
2. Using trigonometric identities, $\frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = (\cot \theta)^2$.
3. Substitute $\cot \theta = \frac{7}{8}$ into the expression to get $\left(\frac{7}{8}\right)^2 = \frac{49}{64}$.

**Answer:** $\frac{49}{64}$

> Common mistake: Squaring the numerator and denominator incorrectly or not simplifying the expression using identities first.

### Question 28

*3 marks · Short answer*

(A) If $x = h + a \cos \theta, y = k + b \sin \theta$, then prove that : $\left(\frac{x - h}{a}\right)^2 + \left(\frac{y - k}{b}\right)^2 = 1$

**Solution**

1. Given equations are $x = h + a \cos \theta$ and $y = k + b \n \theta$.
2. Rearrange the equations to find $\cos \theta$ and $\sin \theta$: $\frac{x - h}{a} = \cos \theta$ and $\frac{y - k}{b} = \sin \theta$.
3. Square both equations: $\left(\frac{x - h}{a}\right)^2 = \cos^2 \theta$ and $\left(\frac{y - k}{b}\right)^2 = \sin^2 \theta$.
4. Add the two squared equations: $\left(\frac{x - h}{a}\right)^2 + \left(\frac{y - k}{b}\right)^2 = \cos^2 \theta + \sin^2 \theta$.
5. Since $\cos^2 \theta + \sin^2 \theta = 1$, we get $\left(\frac{x - h}{a}\right)^2 + \left(\frac{y - k}{b}\right)^2 = 1$.
6. Hence proved.

**Answer:** Hence proved.

> Common mistake: Algebraic errors while squaring the fractions.

### Question 28 (OR)

*3 marks · Short answer*

(B) Prove that : $\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2 \csc A$

**Solution**

1. Consider the LHS: $\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A}$.
2. Take $\tan A$ common from both terms: $\tan A \left(\frac{1}{1 + \sec A} - \frac{1}{1 - \sec A}\right)$.
3. Take the LCM inside the bracket: $\tan A \left[\frac{(1 - \sec A) - (1 + \sec A)}{(1 + \sec A)(1 - \sec A)}\right]$.
4. Simplify the numerator and denominator: $\tan A \left[\frac{-2 \sec A}{1 - \sec^2 A}\right]$.
5. Use the identity $1 - \sec^2 A = -\tan^2 A$: $\tan A \left[\frac{-2 \sec A}{-\tan^2 A}\right]$.
6. Cancel $-\tan A$ from numerator and denominator: $\frac{2 \sec A}{\tan A}$.
7. Convert into sine and cosine: $\frac{2 / \cos A}{\sin A / \cos A} = \frac{2}{\sin A} = 2 \csc A$.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Sign errors while simplifying the numerator.

## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions

### Question 10

*1 mark · MCQ*

Which of the following statements is false ?

- $tan~45^{\circ}=cot~45^{\circ}$
- $sin~90^{\circ}=tan~45^{\circ}$
- $sin~30^{\circ}=cos~30^{\circ}$
- $sin~45^{\circ}=cos~45^{\circ}$

**Solution**

1. We know that $\sin 30^{\circ} = \frac{1}{2}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.
2. Clearly, $\sin 30^{\circ} \neq \cos 30^{\circ}$, making the statement in option (C) false.

**Answer:** (C) $sin~30^{\circ}=cos~30^{\circ}$

> Common mistake: Confusing standard trigonometric table values for 30 degrees.

### Question 11

*1 mark · MCQ*

The value of $(tan^{2}A-\frac{1}{cos^{2}A})$ is:

- more than 1
- 1
- 0
- -1

**Solution**

1. We are given the expression $\tan^{2}A - \frac{1}{\cos^{2}A}$.
2. Since $\frac{1}{\cos^{2}A} = \sec^{2}A$, the expression can be written as $\tan^{2}A - \sec^{2}A$.
3. Using the trigonometric identity $\sec^{2}A - \tan^{2}A = 1$, it follows that $\tan^{2}A - \sec^{2}A = -1$.

**Answer:** (D) -1

> Common mistake: Writing 1 instead of -1 by confusing the order of terms in the identity.

### Question 24 (a)

*2 marks · Very short answer*

Find the values of A and B $(0\le A<90^{\circ}$, $0\le B<90^{\circ})$, if tan (A + B) = 1 and $tan(A-B)=\frac{1}{\sqrt{3}}$

**Solution**

1. We are given $\tan(A + B) = 1$. Since $\tan 45^\circ = 1$, we get $A + B = 45^\circ$.
2. We are also given $\tan(A - B) = \frac{1}{\sqrt{3}}$. Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we get $A - B = 30^\circ$.
3. Adding the two equations: $(A + B) + (A - B) = 45^\circ + 30^\circ$, which gives $2A = 75^\circ$, so $A = 37.5^\circ$.
4. Substituting the value of $A$ in $A + B = 45^\circ$, we get $37.5^\circ + B = 45^\circ$, so $B = 7.5^\circ$.

**Answer:** $A = 37.5^\circ$ and $B = 7.5^\circ$

> Common mistake: Confusing standard angle values for tangent, such as using $30^\circ$ instead of $45^\circ$ for 1.

### Question 24 (OR)

*2 marks · Proof*

Prove that tan $45^{\circ}$ = 1 geometrically.

**Solution**

1. Consider a right-angled triangle $ABC$ right-angled at $B$, where $\angle A = 45^{\circ}$.
2. Since the sum of angles in a triangle is $180^{\circ}$, the third angle $\angle C = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$.
3. Since base angles are equal ($\angle A = \angle C$), the opposite sides are equal, so $AB = BC$.
4. By definition, $\tan 45^{\circ} = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB}$.
5. Since $AB = BC$, $\tan 45^{\circ} = \frac{AB}{AB} = 1$. Hence proved.

**Answer:** $\tan 45^{\circ} = 1$ proved geometrically using an isosceles right-angled triangle.

> Common mistake: Not mentioning that sides are equal because base angles are equal.

### Question 30

*3 marks · Proof*

Prove that : $\frac{1+cot^{2}A}{1+tan^{2}A}=(\frac{1-cot~A}{1-tan~A})^{2}$

**Solution**

1. Consider the left-hand side (LHS): $\frac{1+\cot^{2}A}{1+\tan^{2}A}$.
2. Use the trigonometric identities $1 + \cot^2 A = \csc^2 A$ and $1 + \tan^2 A = \sec^2 A$ to rewrite the expression as $\frac{\csc^{2}A}{\sec^{2}A}$.
3. Express cosecant and secant in terms of sine and cosine: $\csc^2 A = \frac{1}{\sin^2 A}$ and $\sec^2 A = \frac{1}{\cos^2 A}$.
4. Simplify the fraction to get $\frac{\cos^{2}A}{\sin^{2}A} = \cot^{2}A$.
5. Now consider the right-hand side (RHS): $(\frac{1-\cot A}{1-\tan A})^{2}$.
6. Express $\cot A$ as $\frac{\cos A}{\sin A}$ and $\tan A$ as $\frac{\sin A}{\cos A}$ inside the bracket.
7. Simplify the numerator to $\frac{\sin A - \cos A}{\sin A}$ and the denominator to $\frac{\cos A - \sin A}{\cos A}$.
8. Square the expression to obtain $(\frac{\sin A - \cos A}{\sin A} \times \frac{\cos A}{\cos A - \sin A})^{2} = (-\frac{\cos A}{\sin A})^{2} = \cot^{2}A$.
9. Since LHS = RHS, the given identity is proved. Hence proved.

**Answer:** Hence proved that $\frac{1+\cot^{2}A}{1+\tan^{2}A}=(\frac{1-cot~A}{1-tan~A})^{2}$.

> Common mistake: Wrong substitution of trigonometric identities or algebraic errors while squaring the negative term.

## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions

### Question 4

*1 mark · MCQ*

If $\theta$ is an acute angle and $7 + 4 \sin \theta = 9$, then the value of $\theta$ is :

- 90°
- 30°
- 45°
- 60°

**Solution**

1. Given the equation $7 + 4 \sin \theta = 9$, subtract $7$ from both sides to get $4 \sin \theta = 2$.
2. Divide by $4$ to get $\sin \theta = \frac{2}{4} = \frac{1}{2}$.
3. Since $\sin 30^\circ = \frac{1}{2}$ and $\theta$ is an acute angle, $\theta = 30^\circ$.

**Answer:** (b) 30°

> Common mistake: Forgetting standard trigonometric table values and guessing 45°.

### Question 5

*1 mark · MCQ*

The value of $\tan^2 \theta - \left(\frac{1}{\cos \theta} \times \sec \theta\right)$ is :

- 1
- 0
- -1
- 2

**Solution**

1. Consider the expression $\tan^2 \theta - \left(\frac{1}{\cos \theta} \times \sec \theta\right)$.
2. Rewrite $\frac{1}{\cos \theta}$ as $\sec \theta$, so the product becomes $\sec \theta \times \sec \theta = \sec^2 \theta$.
3. The expression simplifies to $\tan^2 \theta - \sec^2 \theta$, which equals $-1$ using the identity $\sec^2 \theta - \tan^2 \theta = 1$.

**Answer:** (c) -1

> Common mistake: Using $\tan^2 \theta - \sec^2 \theta = 1$ instead of $\sec^2 \theta - \tan^2 \theta = 1$, leading to $+1$.

### Question 21

*2 marks · Very short answer*

If $x \cos 60° + y \cos 0° + \sin 30° - \cot 45° = 5$, then find the value of $x + 2y$.

**Solution**

1. Substitute the standard trigonometric values: $\cos 60^\circ = \frac{1}{2}$, $\cos 0^\circ = 1$, $\sin 30^\circ = \frac{1}{2}$, and $\cot 45^\circ = 1$ into the given equation.
2. We get $x\left(\frac{1}{2}\right) + y(1) + \frac{1}{2} - 1 = 5$, which simplifies to $\frac{x}{2} + y - \frac{1}{2} = 5$.
3. Multiply the entire equation by $2$ to clear the fractions: $x + 2y - 1 = 10$, so $x + 2y = 11$.

**Answer:** 11

> Common mistake: Incorrect substitution of trigonometric table values like mixing up sin 30 and cos 60.

### Question 21 (OR)

*2 marks · Very short answer*

Evaluate : $\frac{\tan^2 60°}{\sin^2 60° + \cos^2 30°}$

**Solution**

1. Recall the values: $\tan 60^\circ = \sqrt{3}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, and $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
2. Substitute these values into the expression to get $\frac{(\sqrt{3})^2}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}$.
3. Evaluate the numerator and denominator: $\frac{3}{\frac{3}{4} + \frac{3}{4}} = \frac{3}{\frac{6}{4}} = \frac{3}{\frac{3}{2}} = 2$.

**Answer:** 2

> Common mistake: Squaring the trigonometric ratios incorrectly.

### Question 27

*3 marks · Proof*

Prove that : $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$

**Solution**

1. LHS = $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$
2. Express $\tan \theta$ as $\frac{\sin \theta}{\cos \theta}$ and $\cot \theta$ as $\frac{\cos \theta}{\sin \theta}$.
3. LHS = $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$
4. Taking the LCM, LHS = $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
5. Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, numerator becomes $(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)$.
6. Cancelling $(\sin \theta - \cos \theta)$ and simplifying gives $\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 = 1 + \sec \theta \csc \theta = \text{RHS}$.

**Answer:** Hence proved.

> Common mistake: Algebraic sign errors when converting $\cos \theta - \sin \theta$ to $\sin \theta - \cos \theta$.

### Question 27 (OR)

*3 marks · Proof*

Prove that : $\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2 \sin^2 A - 1}$

**Solution**

1. LHS = $\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A}$
2. Take the LCM of the denominators, which is $(\sin A - \cos A)(\sin A + \cos A) = \sin^2 A - \cos^2 A$.
3. LHS = $\frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{\sin^2 A - \cos^2 A}$
4. Expand the numerators using $(x \pm y)^2$: $\frac{(\sin^2 A + \cos^2 A + 2\sin A \cos A) + (\sin^2 A + \cos^2 A - 2\sin A \cos A)}{\sin^2 A - \cos^2 A}$
5. Simplify the numerator to get $\frac{1 + 1}{\sin^2 A - \cos^2 A} = \frac{2}{\sin^2 A - \cos^2 A}$
6. Substitute $\cos^2 A = 1 - \sin^2 A$ in the denominator: $\frac{2}{\sin^2 A - (1 - \sin^2 A)} = \frac{2}{2\sin^2 A - 1} = \text{RHS}$.

**Answer:** Hence proved.

> Common mistake: Incorrect expansion of binomial squares or incorrect substitution of trigonometric identities in the denominator.

## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions

### Question 4

*1 mark · MCQ*

If $\sin \theta = \frac{1}{3}$, then $\sec \theta$ is equal to :

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

**Solution**

1. We know that $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.
2. Since $\sec \theta = \frac{1}{\cos \theta}$, we have $\sec \theta = \frac{3}{2\sqrt{2}}$.

**Answer:** (b) $\frac{3}{2\sqrt{2}}$

> Common mistake: Inverting $\sin \theta$ directly to find $\sec \theta$.

### Question 9

*1 mark · MCQ*

For what value of $\theta$, $\sin^2 \theta + \sin \theta + \cos^2 \theta$ is equal to $2$ ?

- $45^\circ$
- $0^\circ$
- $90^\circ$
- $30^\circ$

**Solution**

1. We are given the expression $\sin^2 \theta + \sin \theta + \cos^2 \theta = 2$. Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, we get $1 + \sin \theta = 2$.
2. This gives $\sin \theta = 1$, which means $\theta = 90^\circ$.

**Answer:** (c) $90^\circ$

> Common mistake: Forgetting to apply the Pythagorean trigonometric identity first.

### Question 24

*2 marks · Short answer*

Evaluate : $\sin A \cos B + \cos A \sin B$; if $A = 30^\circ$ and $B = 45^\circ$.

**Solution**

1. Given expression is $\sin A \cos B + \cos A \sin B$ with $A = 30^\circ$ and $B = 45^\circ$.
2. Substitute the values into the expression: $\sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ$.
3. Substitute standard trigonometric values: $\left(\frac{1}{2} \times \frac{1}{\sqrt{2}}\right) + \left(\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}\right)$.
4. Simplify the products: $\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}$.
5. Combine the terms over the common denominator: $\frac{1 + \sqrt{3}}{2\sqrt{2}}$.

**Answer:** $\frac{\sqrt{3} + 1}{2\sqrt{2}}$

> Common mistake: Substitution errors for trigonometric values of standard angles.

### Question 31

*3 marks · Proof*

Prove that : $(\cot \theta - \operatorname{cosec} \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$.

**Solution**

1. Consider the Left Hand Side (LHS): $(\cot \theta - \operatorname{cosec} \theta)^2$.
2. Express $\cot \theta$ and $\operatorname{cosec} \theta$ in terms of $\sin \theta$ and $\cos \theta$: $\left(\frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)^2$.
3. Combine the terms inside the bracket: $\left(\frac{\cos \theta - 1}{\sin \theta}\right)^2$.
4. Square the numerator and the denominator separately: $\frac{(\cos \theta - 1)^2}{\sin^2 \theta}$.
5. Rewrite the denominator using $\sin^2 \theta = 1 - \cos^2 \theta$: $\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$.
6. Factor the denominator as $(1 - \cos \theta)(1 + \cos \theta)$ and cancel the common term $(1 - \cos \theta)$: $\frac{1 - \cos \theta}{1 + \cos \theta}$.
7. Hence proved.

**Answer:** Hence proved.

> Common mistake: Making sign errors while writing $(\cos \theta - 1)^2$ as $(1 - \cos \theta)^2$ or failing to apply the algebraic identity for the denominator.

## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions

### Question 7

*1 mark · MCQ*

If $\sec \theta - \tan \theta = m$, then the value of $\sec \theta + \tan \theta$ is :

- $1 - \frac{1}{m}$
- $m^2 - 1$
- $\frac{1}{m}$
- $-m$

**Solution**

1. We know the identity $\sec^2 \theta - \tan^2 \theta = 1$, which can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
2. Substituting the given value $m(\sec \theta + \tan \theta) = 1$, we get $\sec \theta + \tan \theta = \frac{1}{m}$.

**Answer:** (c) $\frac{1}{m}$

> Common mistake: Students often forget the fundamental identity $\sec^2 \theta - \tan^2 \theta = 1$ and try to solve using triangle ratios.

### Question 12

*1 mark · MCQ*

If $\cos(\alpha + \beta) = 0$, then value of $\cos\left(\frac{\alpha + \beta}{2}\right)$ is equal to :

- $\frac{1}{\sqrt{2}}$
- $\frac{1}{2}$
- $0$
- $\sqrt{2}$

**Solution**

1. Given $\cos(\alpha + \beta) = 0$. Since $\cos 90^\circ = 0$, we have $\alpha + \beta = 90^\circ$.
2. We need to find $\cos\left(\frac{\alpha + \beta}{2}\right) = \cos\left(\frac{90^\circ}{2}\right) = \cos 45^\circ$.
3. The value of $\cos 45^\circ$ is $\frac{1}{\sqrt{2}}$.

**Answer:** (a) $\frac{1}{\sqrt{2}}$

> Common mistake: Students incorrectly substitute $\alpha + \beta = 0$ instead of $90^\circ$.

### Question 23

*2 marks · Very short answer*

Evaluate : $2\sqrt{2} \cos 45^\circ \sin 30^\circ + 2\sqrt{3} \cos 30^\circ$

**Solution**

1. Substitute standard trigonometric values: $\cos 45^\circ = \frac{1}{\sqrt{2}}$, $\sin 30^\circ = \frac{1}{2}$, and $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
2. Expression becomes: $2\sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2} + 2\sqrt{3} \times \frac{\sqrt{3}}{2}$.
3. Simplify the terms: $1 + 3 = 4$.

**Answer:** $4$

> Common mistake: Incorrect substitution of trigonometric table values.

### Question 23 (OR)

*2 marks · Very short answer*

If $A = 60^\circ$ and $B = 30^\circ$, verify that : $\sin(A + B) = \sin A \cos B + \cos A \sin B$

**Solution**

1. LHS = $\sin(A + B) = \sin(60^\circ + 30^\circ) = \sin 90^\circ = 1$.
2. RHS = $\sin A \cos B + \cos A \sin B = \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ$.
3. Substitute values: $\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1$.
4. Since LHS = RHS, the identity is verified.

**Answer:** LHS = RHS = 1, hence verified.

> Common mistake: Errors in substituting trigonometric values for specific angles.

### Question 28

*3 marks · Proof*

Prove that : $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$

**Solution**

1. Consider the LHS: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$.
2. Express $\tan \theta$ as $\frac{\sin \theta}{\cos \theta}$ and $\cot \theta$ as $\frac{\cos \theta}{\sin \theta}$.
3. Substitute these into the expression: $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$.
4. Simplify the denominators: $\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$.
5. Rewrite as: $\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)}$.
6. Take the LCM as $\sin \theta \cos \theta (\sin \theta - \cos \theta)$ to get $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$.
7. Use the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ in the numerator: $\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$.
8. Cancel $(\sin \theta - \cos \theta)$ and substitute $\sin^2 \theta + \cos^2 \theta = 1$ to get $\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$.
9. Split the fraction: $\frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} = \sec \theta \csc \theta + 1 = 1 + \sec \theta \csc \theta = \text{RHS}$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Sign errors while changing $\cos \theta - \sin \theta$ to $\sin \theta - \cos \theta$.

## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions

### Question 2

*1 mark · MCQ*

If $2 \cos\theta = 1$, then the value of $\theta$ is

- $45^{\circ}$
- $60^{\circ}$
- $30^{\circ}$
- $90^{\circ}$

**Solution**

1. Given $2 \cos\theta = 1$, we get $\cos\theta = \frac{1}{2}$.
2. Since $\cos 60^{\circ} = \frac{1}{2}$, the value of $\theta$ is $60^{\circ}$.

**Answer:** (b) $60^{\circ}$

> Common mistake: Confusing $\cos 60^{\circ}$ with $\sin 60^{\circ}$ or $\cos 30^{\circ}$.

### Question 22

*2 marks · Short answer*

Evaluate : $\tan^2 60^{\circ} - 2 \operatorname{cosec}^2 30^{\circ} - 2 \tan^2 30^{\circ}$.

**Solution**

1. Substitute the standard trigonometric values: $\tan 60^\circ = \sqrt{3}$, $\operatorname{cosec} 30^\circ = 2$, and $\tan 30^\circ = \frac{1}{\sqrt{3}}$.
2. Write the expression with substituted values: $(\sqrt{3})^2 - 2(2)^2 - 2\left(\frac{1}{\sqrt{3}}\right)^2$.
3. Simplify each term: $3 - 2(4) - 2\left(\frac{1}{3}\right) = 3 - 8 - \frac{2}{3}$.
4. Combine the terms to get $-5 - \frac{2}{3} = \frac{-15 - 2}{3} = \frac{-17}{3}$.

**Answer:** $-\frac{17}{3}$

> Common mistake: Squaring the trigonometric ratio incorrectly or making arithmetic sign errors.

### Question 26

*3 marks · Proof*

Prove that $\sec\theta (1 - \sin\theta) (\sec\theta + \tan\theta) = 1$

**Solution**

1. LHS = $\sec\theta (1 - \sin\theta) (\sec\theta + \tan\theta)$
2. $= \frac{1}{\cos\theta} (1 - \sin\theta) \left(\frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\right)$
3. $= \frac{1 - \sin\theta}{\cos\theta} \left(\frac{1 + \sin\theta}{\cos\theta}\right)$
4. $= \frac{1 - \sin^2\theta}{\cos^2\theta}$
5. $= \frac{\cos^2\theta}{\cos^2\theta} = 1 = \text{RHS}$
6. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not converting secant and tangent into sine and cosine terms when stuck.

### Question 26 (OR)

*3 marks · Proof*

Prove that $\frac{1 + \sec\theta}{\sec\theta} = \frac{\sin^2\theta}{1 - \cos\theta}$

**Solution**

1. LHS = $\frac{1 + \sec\theta}{\sec\theta} = \frac{1 + \frac{1}{\cos\theta}}{\frac{1}{\cos\theta}}$
2. $= \frac{\frac{\cos\theta + 1}{\cos\theta}}{\frac{1}{\cos\theta}} = \cos\theta + 1$
3. RHS = $\frac{\sin^2\theta}{1 - \cos\theta} = \frac{1 - \cos^2\theta}{1 - \cos\theta}$
4. $= \frac{(1 - \cos\theta)(1 + \cos\theta)}{1 - \cos\theta} = 1 + \cos\theta$
5. Since LHS = RHS, Hence proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to factorise $1 - \cos^2\theta$ as $(1 - \cos\theta)(1 + \cos\theta)$.

## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions

### Question 6

*1 mark · MCQ*

If $2 \tan A = 3$, then the value of $\frac{4 \sin A + 3 \cos A}{4 \sin A - 3 \cos A}$ is

- $\frac{7}{\sqrt{13}}$
- $\frac{1}{\sqrt{13}}$
- 3
- does not exist

**Solution**

1. Given $2 \tan A = 3$, which means $\tan A = \frac{3}{2}$.
2. Divide the numerator and denominator of the given expression by $\cos A$ to get $\frac{4 \tan A + 3}{4 \tan A - 3}$.
3. Substitute $\tan A = \frac{3}{2}$ into the expression: $\frac{4\left(\frac{3}{2}\right) + 3}{4\left(\frac{3}{2}\right) - 3}$.
4. Simplify the fraction: $\frac{6 + 3}{6 - 3} = \frac{9}{3} = 3$.

**Answer:** (c) 3

> Common mistake: Trying to construct a right triangle with sides 3 and 2 and applying Pythagoras theorem incorrectly for sin A and cos A without dividing by cos A.

### Question 9

*1 mark · MCQ*

$\left[\frac{3}{4} \tan^2 30^{\circ} - \sec^2 45^{\circ} + \sin^2 60^{\circ}\right]$ is equal to

- -1
- $\frac{5}{6}$
- $\frac{-3}{2}$
- $\frac{1}{6}$

**Solution**

1. Substitute the standard trigonometric values: $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$, $\sec 45^{\circ} = \sqrt{2}$, and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.
2. Evaluate the expression: $\frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)^2 - (\sqrt{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\left(\frac{1}{3}\right) - 2 + \frac{3}{4} = \frac{1}{4} - 2 + \frac{3}{4} = 1 - 2 = -1$.

**Answer:** (a) -1

> Common mistake: Arithmetic errors while combining fractions and whole numbers.

### Question 23

*2 marks · Short answer*

If $\sin \theta + \cos \theta = \sqrt{3}$, then find the value of $\sin \theta \cdot \cos \theta$.

**Solution**

1. Given $\sin \theta + \cos \theta = \sqrt{3}$.
2. Squaring both sides, $(\sin \theta + \cos \theta)^2 = (\sqrt{3})^2$.
3. $\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3$.
4. Using $\sin^2 \theta + \cos^2 \theta = 1$, we get $1 + 2 \sin \theta \cos \theta = 3$.
5. $2 \sin \theta \cos \theta = 2$, which gives $\sin \theta \cos \theta = 1$.

**Answer:** $1$

> Common mistake: Forgetting to square the right-hand side properly or missing the middle term $2 \sin \theta \cos \theta$.

### Question 23 (OR)

*2 marks · Short answer*

If $\sin \alpha = \frac{1}{\sqrt{2}}$ and $\cot \beta = \sqrt{3}$, then find the value of $\csc \alpha + \csc \beta$.

**Solution**

1. Given $\sin \alpha = \frac{1}{\sqrt{2}}$, which means $\alpha = 45^\circ$, so $\csc \alpha = \sqrt{2}$.
2. Given $\cot \beta = \sqrt{3}$, which means $\beta = 30^\circ$, so $\csc \beta = 2$.
3. Therefore, $\csc \alpha + \csc \beta = \sqrt{2} + 2$.

**Answer:** $\sqrt{2} + 2$

> Common mistake: Mixing up trigonometric ratios like sine with cosecant or tangent with cotangent.

### Question 30

*3 marks · Proof*

Prove that : $\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}$

**Solution**

1. Consider the LHS: $\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1}$.
2. Using the trigonometric identity $\sec^2 \theta - \tan^2 \theta = 1$, we can write $1 = \sec^2 \theta - \tan^2 \theta$.
3. Substitute this in the numerator: $\frac{(\tan \theta + \sec \theta) - (\sec^2 \theta - \tan^2 \theta)}{\tan \theta - \sec \theta + 1}$.
4. Factor the numerator: $\frac{(\tan \theta + \sec \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\tan \theta - \sec \theta + 1}$.
5. Take $(\tan \theta + \sec \theta)$ common from the numerator: $\frac{(\tan \theta + \sec \theta)(1 - (\sec \theta - \tan \theta))}{\tan \theta - \sec \theta + 1}$.
6. Simplify the second bracket in the numerator to get $1 - \sec \theta + \tan \theta$, which cancels out with the denominator $\tan \theta - \sec \theta + 1$.
7. We are left with $\tan \theta + \sec \theta$, which can be written as $\frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = \frac{1 + \sin \theta}{\cos \theta}$.
8. Hence proved.

**Answer:** Hence proved that $\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}$.

> Common mistake: Applying the identity $1 = \sec^2 \theta - \tan^2 \theta$ in the denominator instead of the numerator.

## Related pages

- [Introduction to Trigonometry: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/introduction-to-trigonometry/ncert-solutions)
- [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers)

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