This page provides comprehensive Introduction to Trigonometry – Exercise 8.3. Get step-by-step NCERT solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3. Learn Trigonometric Identities and how to prove them.
Trigonometric Identities
Remember these three key identities:
1. $\sin^2 A + \cos^2 A = 1$
2. $1 + \tan^2 A = \sec^2 A$
3. $1 + \cot^2 A = \text{cosec}^2 A$
1. Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.
(i) For $\tan A$:
$\tan A = \frac{1}{\cot A}$
(ii) For $\sec A$:
$\sec^2 A = 1 + \tan^2 A = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A}$
$\Rightarrow \sec A = \frac{\sqrt{\cot^2 A + 1}}{\cot A}$
(iii) For $\sin A$:
$\sin A = \frac{1}{\text{cosec} A} = \frac{1}{\sqrt{1+\cot^2 A}}$
2. Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.
1. $\cos A = \frac{1}{\sec A}$
2. $\sin^2 A = 1 - \cos^2 A = 1 - \frac{1}{\sec^2 A} = \frac{\sec^2 A - 1}{\sec^2 A}$
$\Rightarrow \sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$
3. $\tan^2 A = \sec^2 A - 1 \Rightarrow \tan A = \sqrt{\sec^2 A - 1}$
4. $\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}$
5. $\text{cosec} A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$
3. Choose the correct option. Justify your choice.
(i) $9 \sec^2 A - 9 \tan^2 A =$
(A) 1 (B) 9 (C) 8 (D) 0
$= 9(\sec^2 A - \tan^2 A)$
Using identity $\sec^2 A - \tan^2 A = 1$:
$= 9(1) = 9$
Option (B) is correct.
(ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta - \text{cosec} \theta) =$
(A) 0 (B) 1 (C) 2 (D) -1
Convert to sin/cos:
$= (1 + \frac{\sin}{\cos} + \frac{1}{\cos})(1 + \frac{\cos}{\sin} - \frac{1}{\sin})$
$= (\frac{\cos+\sin+1}{\cos})(\frac{\sin+\cos-1}{\sin})$
$= \frac{(\sin+\cos)^2 - (1)^2}{\sin \cos} = \frac{\sin^2+\cos^2+2\sin\cos - 1}{\sin \cos}$
$= \frac{1 + 2\sin\cos - 1}{\sin \cos} = \frac{2\sin\cos}{\sin\cos} = 2$
Option (C) is correct.
(iii) $(\sec A + \tan A) (1 - \sin A) =$
(A) $\sec A$ (B) $\sin A$ (C) $\text{cosec} A$ (D) $\cos A$
$= (\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A)$
$= (\frac{1+\sin A}{\cos A})(1 - \sin A) = \frac{1 - \sin^2 A}{\cos A}$
$= \frac{\cos^2 A}{\cos A} = \cos A$
Option (D) is correct.
(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$
(A) $\sec^2 A$ (B) -1 (C) $\cot^2 A$ (D) $\tan^2 A$
$= \frac{\sec^2 A}{\text{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$
Option (D) is correct.
4. Prove the following identities:
(i) $(\text{cosec} \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
LHS $= (\frac{1}{\sin} - \frac{\cos}{\sin})^2 = (\frac{1-\cos}{\sin})^2 = \frac{(1-\cos)^2}{\sin^2}$
$= \frac{(1-\cos)^2}{1-\cos^2} = \frac{(1-\cos)(1-\cos)}{(1-\cos)(1+\cos)} = \frac{1-\cos}{1+\cos} =$ RHS
(ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$
LHS $= \frac{\cos^2 A + (1+\sin A)^2}{\cos A(1+\sin A)} = \frac{\cos^2 A + 1 + \sin^2 A + 2\sin A}{\cos A(1+\sin A)}$
$= \frac{2 + 2\sin A}{\cos A(1+\sin A)} = \frac{2(1+\sin A)}{\cos A(1+\sin A)} = \frac{2}{\cos A} = 2\sec A =$ RHS
(iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec} \theta$
Convert to sin/cos. Numerator becomes $\sin^2/(\cos(\sin-\cos)) - \cos^2/(\sin(\sin-\cos))$
$= \frac{\sin^3 - \cos^3}{\sin\cos(\sin-\cos)} = \frac{(\sin-\cos)(\sin^2+\cos^2+\sin\cos)}{\sin\cos(\sin-\cos)}$
$= \frac{1+\sin\cos}{\sin\cos} = \frac{1}{\sin\cos} + 1 = 1 + \sec\theta\text{cosec}\theta =$ RHS
(iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A}$
LHS $= \frac{1 + 1/\cos}{1/\cos} = \cos A + 1$
RHS $= \frac{1-\cos^2 A}{1-\cos A} = \frac{(1-\cos)(1+\cos)}{1-\cos} = 1 + \cos A$
LHS = RHS
(v) $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec} A + \cot A$
Divide num and den by $\sin A$:
$= \frac{\cot A - 1 + \text{cosec} A}{\cot A + 1 - \text{cosec} A}$. Replace $1$ in num with $\text{cosec}^2 - \cot^2$.
$= \frac{(\cot + \text{cosec}) - (\text{cosec} - \cot)(\text{cosec} + \cot)}{\cot - \text{cosec} + 1}$
$= \frac{(\text{cosec} + \cot)(1 - \text{cosec} + \cot)}{1 - \text{cosec} + \cot} = \text{cosec} A + \cot A =$ RHS
(vi) $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$
Rationalize inside root: $\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)} = \frac{(1+\sin A)^2}{1-\sin^2 A}$
$= \frac{(1+\sin A)^2}{\cos^2 A}$. Take square root: $\frac{1+\sin A}{\cos A} = \sec A + \tan A =$ RHS
(vii) $\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta} = \tan \theta$
LHS $= \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)}$
Use $1-2\sin^2 = \cos 2\theta$ and $2\cos^2 - 1 = \cos 2\theta$.
$= \frac{\sin \theta (\cos 2\theta)}{\cos \theta (\cos 2\theta)} = \tan \theta =$ RHS
(viii) $(\sin A + \text{cosec} A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
Expand: $\sin^2 + \text{cosec}^2 + 2 + \cos^2 + \sec^2 + 2$
$(\sin^2+\cos^2) + (\text{cosec}^2) + (\sec^2) + 4 = 1 + (1+\cot^2) + (1+\tan^2) + 4$
$= 7 + \tan^2 A + \cot^2 A =$ RHS
(ix) $(\text{cosec} A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$
LHS $= (\frac{1-\sin^2}{\sin})(\frac{1-\cos^2}{\cos}) = \frac{\cos^2}{\sin} \cdot \frac{\sin^2}{\cos} = \sin A \cos A$
RHS $= \frac{1}{\frac{\sin}{\cos} + \frac{\cos}{\sin}} = \frac{1}{\frac{\sin^2+\cos^2}{\sin\cos}} = \sin A \cos A$
LHS = RHS
(x) $(\frac{1 + \tan^2 A}{1 + \cot^2 A}) = (\frac{1 - \tan A}{1 - \cot A})^2 = \tan^2 A$
Part 1: $\frac{\sec^2}{\text{cosec}^2} = \tan^2 A$.
Part 2: $(\frac{1-\tan}{1-1/\tan})^2 = (\frac{1-\tan}{(\tan-1)/\tan})^2 = (-\tan)^2 = \tan^2 A$.
All parts equal.