Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 8 Exercise 8.3.
Master the use of Pythagorean Identities ($\sin^2 \theta + \cos^2 \theta = 1$, $\sec^2 \theta = 1 + \tan^2 \theta$, $\csc^2 \theta = 1 + \cot^2 \theta$) and algebraic techniques to solve and prove complex trigonometric identities (Q.4, i-x). Practice expressing all trigonometric ratios in terms of a single ratio like $\cot A$ or $\sec A$ (Q.1, Q.2). Solutions include techniques like multiplying by conjugates, taking common denominators, and factorization (Q.4, v, vi, vii). Also covers simplification of MCQ-type problems based on fundamental identities (Q.3). Essential for developing proficiency in algebraic trigonometry.
This exercise focuses on using the fundamental trigonometric identities to express ratios in terms of one another, simplify expressions, and prove identities.
The key identities used are:
- $\sin^2 \theta + \cos^2 \theta = 1$
- $\sec^2 \theta = 1 + \tan^2 \theta$
- $\csc^2 \theta = 1 + \cot^2 \theta$
1. Express $\sin A, \sec A$, and $\tan A$ in terms of $\cot A$.
- $\tan A$ in terms of $\cot A$:$$\tan A = \frac{1}{\cot A}$$
- $\sin A$ in terms of $\cot A$:Start with the identity $\csc^2 A = 1 + \cot^2 A$.$$\csc A = \sqrt{1 + \cot^2 A}$$$$\sin A = \frac{1}{\csc A} = \mathbf{\frac{1}{\sqrt{1 + \cot^2 A}}}$$
- $\sec A$ in terms of $\cot A$:Start with the identity $\sec^2 A = 1 + \tan^2 A$ and substitute $\tan A = 1/\cot A$:$$\sec^2 A = 1 + \left(\frac{1}{\cot A}\right)^2 = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A}$$$$\sec A = \sqrt{\frac{\cot^2 A + 1}{\cot^2 A}} = \mathbf{\frac{\sqrt{1 + \cot^2 A}}{\cot A}}$$
2. Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.
The required ratios are $\sin A, \cos A, \tan A, \csc A, \cot A$.
- $\cos A$ in terms of $\sec A$:$$\cos A = \mathbf{\frac{1}{\sec A}}$$
- $\tan A$ in terms of $\sec A$:Start with the identity $\tan^2 A = \sec^2 A – 1$.$$\tan A = \mathbf{\sqrt{\sec^2 A – 1}}$$
- $\sin A$ in terms of $\sec A$:Start with $\sin A = \tan A \cdot \cos A$.$$\sin A = \left(\sqrt{\sec^2 A – 1}\right) \left(\frac{1}{\sec A}\right) = \mathbf{\frac{\sqrt{\sec^2 A – 1}}{\sec A}}$$
- $\cot A$ in terms of $\sec A$:Start with $\cot A = \frac{1}{\tan A}$.$$\cot A = \mathbf{\frac{1}{\sqrt{\sec^2 A – 1}}}$$
- $\csc A$ in terms of $\sec A$:Start with $\csc A = \frac{1}{\sin A}$.$$\csc A = \frac{1}{\frac{\sqrt{\sec^2 A – 1}}{\sec A}} = \mathbf{\frac{\sec A}{\sqrt{\sec^2 A – 1}}}$$
3. Choose the correct option. Justify your choice.
(i) $9 \sec^2 A – 9 \tan^2 A =$
Factor out 9:
$$9 (\sec^2 A – \tan^2 A)$$
Using the identity $\sec^2 A – \tan^2 A = 1$:
$$9(1) = 9$$
The correct option is (B) 9.
(ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta – \csc \theta) =$
Convert all terms to $\sin \theta$ and $\cos \theta$:
$$LHS = \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right) \left(1 + \frac{\cos \theta}{\sin \theta} – \frac{1}{\sin \theta}\right)$$
$$LHS = \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\frac{\sin \theta + \cos \theta – 1}{\sin \theta}\right)$$
Let $X = \sin \theta + \cos \theta$. The numerator is of the form $(X + 1)(X – 1) = X^2 – 1^2$:
$$LHS = \frac{(\sin \theta + \cos \theta)^2 – 1}{\cos \theta \sin \theta}$$
Expand the square:
$$LHS = \frac{(\sin^2 \theta + \cos^2 \theta) + 2 \sin \theta \cos \theta – 1}{\cos \theta \sin \theta}$$
Since $\sin^2 \theta + \cos^2 \theta = 1$:
$$LHS = \frac{1 + 2 \sin \theta \cos \theta – 1}{\cos \theta \sin \theta} = \frac{2 \sin \theta \cos \theta}{\cos \theta \sin \theta} = 2$$
The correct option is (C) 2.
(iii) $(\sec A + \tan A) (1 – \sin A) =$
Convert to $\sin A$ and $\cos A$:
$$LHS = \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) (1 – \sin A)$$
$$LHS = \left(\frac{1 + \sin A}{\cos A}\right) (1 – \sin A) = \frac{(1 + \sin A)(1 – \sin A)}{\cos A}$$
Using the identity $(1 + \sin A)(1 – \sin A) = 1 – \sin^2 A$:
$$LHS = \frac{1 – \sin^2 A}{\cos A}$$
Using the identity $1 – \sin^2 A = \cos^2 A$:
$$LHS = \frac{\cos^2 A}{\cos A} = \cos A$$
The correct option is (D) $\cos A$.
(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$
Use the identities $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \csc^2 A$:
$$\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A}$$
Convert to $\sin A$ and $\cos A$:
$$\frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$
The correct option is (D) $\tan^2 A$.
4. Prove the following identities
(i) $(\csc \theta – \cot \theta)^2 = \frac{1 – \cos \theta}{1 + \cos \theta}$
$$\text{LHS} = (\csc \theta – \cot \theta)^2$$
Convert to $\sin \theta$ and $\cos \theta$:
$$LHS = \left(\frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta}\right)^2 = \left(\frac{1 – \cos \theta}{\sin \theta}\right)^2$$
$$LHS = \frac{(1 – \cos \theta)^2}{\sin^2 \theta}$$
Use the identity $\sin^2 \theta = 1 – \cos^2 \theta$:
$$LHS = \frac{(1 – \cos \theta)^2}{1 – \cos^2 \theta}$$
Factor the denominator using $a^2 – b^2 = (a – b)(a + b)$:
$$LHS = \frac{(1 – \cos \theta)(1 – \cos \theta)}{(1 – \cos \theta)(1 + \cos \theta)} = \frac{1 – \cos \theta}{1 + \cos \theta} = \text{RHS}$$
(ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$
Take LCM on the LHS:
$$\text{LHS} = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A}$$
Expand the numerator:
$$\text{LHS} = \frac{\cos^2 A + (1 + 2 \sin A + \sin^2 A)}{(1 + \sin A)\cos A}$$
Group $\cos^2 A + \sin^2 A = 1$:
$$\text{LHS} = \frac{(\cos^2 A + \sin^2 A) + 1 + 2 \sin A}{(1 + \sin A)\cos A} = \frac{1 + 1 + 2 \sin A}{(1 + \sin A)\cos A}$$
$$\text{LHS} = \frac{2 + 2 \sin A}{(1 + \sin A)\cos A} = \frac{2(1 + \sin A)}{(1 + \sin A)\cos A}$$
Cancel $(1 + \sin A)$:
$$\text{LHS} = \frac{2}{\cos A} = 2 \left(\frac{1}{\cos A}\right) = 2 \sec A = \text{RHS}$$
(iii) $\frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta} = 1 + \sec \theta \csc \theta$
Convert LHS to $\sin \theta$ and $\cos \theta$:
$$\text{LHS} = \frac{\sin \theta/\cos \theta}{1 – \cos \theta/\sin \theta} + \frac{\cos \theta/\sin \theta}{1 – \sin \theta/\cos \theta}$$
Simplify the denominators:
$$\text{LHS} = \frac{\sin \theta/\cos \theta}{(\sin \theta – \cos \theta)/\sin \theta} + \frac{\cos \theta/\sin \theta}{(\cos \theta – \sin \theta)/\cos \theta}$$
$$\text{LHS} = \frac{\sin \theta}{\cos \theta} \cdot \frac{\sin \theta}{\sin \theta – \cos \theta} + \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\cos \theta – \sin \theta}$$
Note that $(\cos \theta – \sin \theta) = -(\sin \theta – \cos \theta)$. Factor out the common denominator $(\sin \theta – \cos \theta)$:
$$\text{LHS} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta – \cos \theta)} – \frac{\cos^2 \theta}{\sin \theta (\sin \theta – \cos \theta)}$$
Take LCM $\sin \theta \cos \theta (\sin \theta – \cos \theta)$:
$$\text{LHS} = \frac{\sin^3 \theta – \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta – \cos \theta)}$$
Use the identity $a^3 – b^3 = (a – b)(a^2 + ab + b^2)$, where $a=\sin \theta, b=\cos \theta$:
$$\text{LHS} = \frac{(\sin \theta – \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta – \cos \theta)}$$
Cancel $(\sin \theta – \cos \theta)$ and use $\sin^2 \theta + \cos^2 \theta = 1$:
$$\text{LHS} = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$$
Separate the terms:
$$\text{LHS} = \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} = \csc \theta \sec \theta + 1 = \text{RHS}$$
(iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$
Simplify LHS:
$$\text{LHS} = \frac{1 + \sec A}{\sec A} = \frac{1}{\sec A} + \frac{\sec A}{\sec A} = \cos A + 1$$
Simplify RHS:
$$\text{RHS} = \frac{\sin^2 A}{1 – \cos A}$$
Use $\sin^2 A = 1 – \cos^2 A$:
$$\text{RHS} = \frac{1 – \cos^2 A}{1 – \cos A}$$
Factor the numerator:
$$\text{RHS} = \frac{(1 – \cos A)(1 + \cos A)}{1 – \cos A} = 1 + \cos A$$
Since $\text{LHS} = 1 + \cos A$ and $\text{RHS} = 1 + \cos A$, we have $\text{LHS} = \text{RHS}$.
(v) $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \csc A + \cot A$ (Using $\csc^2 A = 1 + \cot^2 A$)
Divide the numerator and denominator of the LHS by $\sin A$:
$$\text{LHS} = \frac{\frac{\cos A}{\sin A} – \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} – \frac{1}{\sin A}} = \frac{\cot A – 1 + \csc A}{\cot A + 1 – \csc A}$$
In the numerator, substitute the identity $\mathbf{1 = \csc^2 A – \cot^2 A}$:
$$\text{LHS} = \frac{\csc A + \cot A – (\csc^2 A – \cot^2 A)}{1 + \cot A – \csc A}$$
Factor the difference of squares: $\csc^2 A – \cot^2 A = (\csc A – \cot A)(\csc A + \cot A)$:
$$\text{LHS} = \frac{(\csc A + \cot A) – (\csc A – \cot A)(\csc A + \cot A)}{1 + \cot A – \csc A}$$
Factor out $(\csc A + \cot A)$ from the numerator:
$$\text{LHS} = \frac{(\csc A + \cot A) [1 – (\csc A – \cot A)]}{1 + \cot A – \csc A}$$
$$\text{LHS} = \frac{(\csc A + \cot A) [1 – \csc A + \cot A]}{1 + \cot A – \csc A}$$
Since $1 + \cot A – \csc A$ is the same in the numerator and denominator, they cancel:
$$\text{LHS} = \csc A + \cot A = \text{RHS}$$
(vi) $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$
Multiply the numerator and denominator inside the square root by the conjugate of the denominator, $(1 + \sin A)$:
$$\text{LHS} = \sqrt{\frac{1 + \sin A}{1 – \sin A} \times \frac{1 + \sin A}{1 + \sin A}} = \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}}$$
Use the identity $1 – \sin^2 A = \cos^2 A$:
$$\text{LHS} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$$
Take the square root:
$$\text{LHS} = \frac{1 + \sin A}{\cos A}$$
Separate the terms:
$$\text{LHS} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS}$$
(vii) $\frac{\sin \theta – 2 \sin^3 \theta}{2 \cos^3 \theta – \cos \theta} = \tan \theta$
Factor out $\sin \theta$ from the numerator and $\cos \theta$ from the denominator:
$$\text{LHS} = \frac{\sin \theta (1 – 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta – 1)}$$
Since $\frac{\sin \theta}{\cos \theta} = \tan \theta$, we just need to show that $\frac{1 – 2 \sin^2 \theta}{2 \cos^2 \theta – 1} = 1$.
Use the identity $\sin^2 \theta = 1 – \cos^2 \theta$ in the numerator:
$$1 – 2 \sin^2 \theta = 1 – 2(1 – \cos^2 \theta) = 1 – 2 + 2 \cos^2 \theta = 2 \cos^2 \theta – 1$$
Substitute this back into the fraction:
$$\text{LHS} = \frac{\sin \theta}{\cos \theta} \cdot \frac{2 \cos^2 \theta – 1}{2 \cos^2 \theta – 1} = \tan \theta \cdot 1 = \tan \theta = \text{RHS}$$
(viii) $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
Expand the squares using $(a+b)^2 = a^2 + 2ab + b^2$:
$$\text{LHS} = (\sin^2 A + 2 \sin A \csc A + \csc^2 A) + (\cos^2 A + 2 \cos A \sec A + \sec^2 A)$$
Simplify terms using $\sin A \csc A = 1$ and $\cos A \sec A = 1$:
$$\text{LHS} = (\sin^2 A + 2(1) + \csc^2 A) + (\cos^2 A + 2(1) + \sec^2 A)$$
Group $\sin^2 A$ and $\cos^2 A$:
$$\text{LHS} = (\sin^2 A + \cos^2 A) + 2 + 2 + \csc^2 A + \sec^2 A$$
Use $\sin^2 A + \cos^2 A = 1$:
$$\text{LHS} = 1 + 4 + \csc^2 A + \sec^2 A$$
Use the identities $\csc^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$:
$$\text{LHS} = 5 + (1 + \cot^2 A) + (1 + \tan^2 A)$$
$$\text{LHS} = 7 + \tan^2 A + \cot^2 A = \text{RHS}$$
(ix) $(\csc A – \sin A)(\sec A – \cos A) = \frac{1}{\tan A + \cot A}$
Simplify LHS:
Convert to $\sin A$ and $\cos A$:
$$\text{LHS} = \left(\frac{1}{\sin A} – \sin A\right) \left(\frac{1}{\cos A} – \cos A\right)$$
$$\text{LHS} = \left(\frac{1 – \sin^2 A}{\sin A}\right) \left(\frac{1 – \cos^2 A}{\cos A}\right)$$
Use $1 – \sin^2 A = \cos^2 A$ and $1 – \cos^2 A = \sin^2 A$:
$$\text{LHS} = \frac{\cos^2 A}{\sin A} \cdot \frac{\sin^2 A}{\cos A}$$
Cancel terms:
$$\text{LHS} = \cos A \sin A$$
Simplify RHS:
Convert to $\sin A$ and $\cos A$:
$$\text{RHS} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$$
Take LCM in the denominator:
$$\text{RHS} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\cos A \sin A}}$$
Use $\sin^2 A + \cos^2 A = 1$:
$$\text{RHS} = \frac{1}{\frac{1}{\cos A \sin A}} = \cos A \sin A$$
Since $\text{LHS} = \cos A \sin A$ and $\text{RHS} = \cos A \sin A$, we have $\text{LHS} = \text{RHS}$.
(x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 – \tan A}{1 – \cot A}\right)^2 = \tan^2 A$
This identity requires two proofs.
Part 1: $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A$
$$\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$
(Proved in Q.3, iv).
Part 2: $\left(\frac{1 – \tan A}{1 – \cot A}\right)^2 = \tan^2 A$
Convert $\cot A$ to $1/\tan A$:
$$\text{LHS} = \left(\frac{1 – \tan A}{1 – \frac{1}{\tan A}}\right)^2$$
Simplify the denominator:
$$\text{LHS} = \left(\frac{1 – \tan A}{\frac{\tan A – 1}{\tan A}}\right)^2$$
Since $1 – \tan A = -(\tan A – 1)$:
$$\text{LHS} = \left(\frac{-( \tan A – 1)}{(\tan A – 1)} \cdot \tan A\right)^2$$
Cancel $(\tan A – 1)$:
$$\text{LHS} = (-\tan A)^2 = \tan^2 A$$
Since both parts equal $\tan^2 A$, the identity is proved.
Last Updated on November 28, 2025 by Aman Singh
