Exercise 8.3 Practice
Trigonometric Identities (New Questions)
Key Identities for this Exercise
- 1. $\sin^2 A + \cos^2 A = 1$
- 2. $1 + \tan^2 A = \sec^2 A$
- 3. $1 + \cot^2 A = \text{cosec}^2 A$
- $\tan A = \frac{\sin A}{\cos A}$
- $\cot A = \frac{\cos A}{\sin A}$
Q1: Ratio Conversion
Express the trigonometric ratios $\sin A$, $\sec A$, and $\tan A$ in terms of $\text{cosec } A$.
1. For $\sin A$:
$\sin A = \frac{1}{\text{cosec } A}$
$\sin A = \frac{1}{\text{cosec } A}$
2. For $\tan A$:
Using identity $1 + \cot^2 A = \text{cosec}^2 A$:
$\cot^2 A = \text{cosec}^2 A - 1$
$\tan^2 A = \frac{1}{\cot^2 A} = \frac{1}{\text{cosec}^2 A - 1}$
$\tan A = \frac{1}{\sqrt{\text{cosec}^2 A - 1}}$
Using identity $1 + \cot^2 A = \text{cosec}^2 A$:
$\cot^2 A = \text{cosec}^2 A - 1$
$\tan^2 A = \frac{1}{\cot^2 A} = \frac{1}{\text{cosec}^2 A - 1}$
$\tan A = \frac{1}{\sqrt{\text{cosec}^2 A - 1}}$
3. For $\sec A$:
$\sec^2 A = 1 + \tan^2 A = 1 + \frac{1}{\text{cosec}^2 A - 1}$
$\sec^2 A = \frac{\text{cosec}^2 A - 1 + 1}{\text{cosec}^2 A - 1} = \frac{\text{cosec}^2 A}{\text{cosec}^2 A - 1}$
$\sec A = \frac{\text{cosec } A}{\sqrt{\text{cosec}^2 A - 1}}$
$\sec^2 A = 1 + \tan^2 A = 1 + \frac{1}{\text{cosec}^2 A - 1}$
$\sec^2 A = \frac{\text{cosec}^2 A - 1 + 1}{\text{cosec}^2 A - 1} = \frac{\text{cosec}^2 A}{\text{cosec}^2 A - 1}$
$\sec A = \frac{\text{cosec } A}{\sqrt{\text{cosec}^2 A - 1}}$
Q2: Express Ratios
Write all the other trigonometric ratios of $\angle A$ in terms of $\cot A$.
1. $\tan A$: $\frac{1}{\cot A}$
2. $\text{cosec } A$: $\text{cosec } A = \sqrt{1 + \cot^2 A}$
3. $\sin A$: $\sin A = \frac{1}{\text{cosec } A} = \frac{1}{\sqrt{1 + \cot^2 A}}$
4. $\cos A$: $\cos A = \sin A \cdot \cot A = \frac{\cot A}{\sqrt{1 + \cot^2 A}}$
5. $\sec A$: $\sec A = \frac{1}{\cos A} = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$
Q3: Multiple Choice
Choose the correct option. Justify your choice.
(i) $4 \cot^2 A - 4 \text{cosec}^2 A =$
(A) 4 (B) -4 (C) 0 (D) 1
(ii) $(\sec \theta + \tan \theta)(1 - \sin \theta) =$
(A) $\sec \theta$ (B) $\sin \theta$ (C) $\text{cosec } \theta$ (D) $\cos \theta$
(iii) $\frac{1 + \cot^2 A}{1 + \tan^2 A} =$
(A) $\sec^2 A$ (B) $\cot^2 A$ (C) $\tan^2 A$ (D) -1
(i) $4 \cot^2 A - 4 \text{cosec}^2 A =$
(A) 4 (B) -4 (C) 0 (D) 1
(ii) $(\sec \theta + \tan \theta)(1 - \sin \theta) =$
(A) $\sec \theta$ (B) $\sin \theta$ (C) $\text{cosec } \theta$ (D) $\cos \theta$
(iii) $\frac{1 + \cot^2 A}{1 + \tan^2 A} =$
(A) $\sec^2 A$ (B) $\cot^2 A$ (C) $\tan^2 A$ (D) -1
(i) $4(\cot^2 A - \text{cosec}^2 A) = 4(-1) = -4$. (Since $\text{cosec}^2 - \cot^2 = 1$). (B)
(ii) $(\frac{1}{\cos} + \frac{\sin}{\cos})(1 - \sin) = \frac{1+\sin}{\cos}(1-\sin) = \frac{1-\sin^2}{\cos} = \frac{\cos^2}{\cos} = \cos \theta$. (D)
(iii) $\frac{\text{cosec}^2 A}{\sec^2 A} = \frac{1/\sin^2 A}{1/\cos^2 A} = \frac{\cos^2 A}{\sin^2 A} = \cot^2 A$. (B)
Q4: Identities Proof
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) $(\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta}$
(ii) $\frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} = 2 \text{cosec } A$
(iii) $\frac{1}{\sec A - \tan A} = \sec A + \tan A$
(iv) $\frac{1 + \text{cosec } A}{\text{cosec } A} = \frac{\cos^2 A}{1 - \sin A}$
(v) $\frac{2 \cos^3 \theta - \cos \theta}{\sin \theta - 2 \sin^3 \theta} = \cot \theta$
(vi) $\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \text{cosec } A - \cot A$
(vii) $\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A$
(viii) $(\sin A + \sec A)^2 + (\cos A + \text{cosec } A)^2 = (1 + \sec A \text{cosec } A)^2$
(ix) $(\sec A - \cos A)(\text{cosec } A - \sin A) = \frac{1}{\tan A + \cot A}$
(x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$
(i) $(\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta}$
(ii) $\frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} = 2 \text{cosec } A$
(iii) $\frac{1}{\sec A - \tan A} = \sec A + \tan A$
(iv) $\frac{1 + \text{cosec } A}{\text{cosec } A} = \frac{\cos^2 A}{1 - \sin A}$
(v) $\frac{2 \cos^3 \theta - \cos \theta}{\sin \theta - 2 \sin^3 \theta} = \cot \theta$
(vi) $\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \text{cosec } A - \cot A$
(vii) $\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A$
(viii) $(\sin A + \sec A)^2 + (\cos A + \text{cosec } A)^2 = (1 + \sec A \text{cosec } A)^2$
(ix) $(\sec A - \cos A)(\text{cosec } A - \sin A) = \frac{1}{\tan A + \cot A}$
(x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$
(i) LHS: $(\frac{1}{\cos} - \frac{\sin}{\cos})^2 = (\frac{1-\sin}{\cos})^2 = \frac{(1-\sin)^2}{1-\sin^2} = \frac{(1-\sin)(1-\sin)}{(1-\sin)(1+\sin)} = \text{RHS}$.
(ii) LHS: $\frac{\sin^2 + (1+\cos)^2}{(1+\cos)\sin} = \frac{\sin^2 + 1 + \cos^2 + 2\cos}{(1+\cos)\sin} = \frac{2 + 2\cos}{(1+\cos)\sin} = \frac{2(1+\cos)}{(1+\cos)\sin} = \frac{2}{\sin} = 2\text{cosec } A$.
(iii) LHS: Rationalize: $\frac{1(\sec + \tan)}{(\sec - \tan)(\sec + \tan)} = \frac{\sec + \tan}{\sec^2 - \tan^2} = \frac{\sec + \tan}{1} = \text{RHS}$.
(iv) LHS: $\frac{1+1/\sin}{1/\sin} = \sin + 1$. RHS: $\frac{1-\sin^2}{1-\sin} = 1+\sin$. LHS = RHS.
(v) LHS: $\frac{\cos(2\cos^2 - 1)}{\sin(1 - 2\sin^2)}$. Since $2\cos^2 - 1 = \cos 2\theta$ and $1 - 2\sin^2 = \cos 2\theta$, they cancel out. Result: $\frac{\cos}{\sin} = \cot \theta$.
(vi) LHS: Multiply num & den inside root by $(1-\cos)$. $\sqrt{\frac{(1-\cos)^2}{1-\cos^2}} = \frac{1-\cos}{\sin} = \text{cosec } - \cot$.
(vii) LHS: Convert to sin/cos. $\frac{\cos}{1-s/c} + \frac{\sin}{1-c/s} = \frac{c^2}{c-s} + \frac{s^2}{s-c} = \frac{c^2 - s^2}{c-s} = \frac{(c-s)(c+s)}{c-s} = \cos + \sin$.
(ix) LHS: $(\frac{1}{c}-c)(\frac{1}{s}-s) = \frac{s^2}{c} \cdot \frac{c^2}{s} = sc$. RHS: $\frac{1}{s/c + c/s} = \frac{1}{1/sc} = sc$. LHS = RHS.
(x) $\frac{\sec^2}{\text{cosec}^2} = \tan^2$. Middle term: $(\frac{1-\tan}{1-1/\tan})^2 = (\frac{1-\tan}{(\tan-1)/\tan})^2 = (-\tan)^2 = \tan^2$.
All Identities Proved.