Choose the correct option: 9 sec² A - 9 tan² A =

The correct option is (B) 9. Justification: 9(sec² A - tan² A) = 9(1) = 9, since sec² A - tan² A = 1.

Choose the correct option: (sec A + tan A) (1 - sin A) =

The correct option is (D) cos A. Justification: (1/cos A + sin A/cos A)(1 - sin A) = [(1 + sin A)(1 - sin A)] / cos A = (1 - sin² A) / cos A = cos² A / cos A = cos A.

Prove the identity: (cosec θ - cot θ)² = (1 - cos θ) / (1 + cos θ)

LHS = (1/sin θ - cos θ/sin θ)² = (1 - cos θ)² / sin² θ = (1 - cos θ)² / (1 - cos² θ) = (1 - cos θ)(1 - cos θ) / [(1 - cos θ)(1 + cos θ)] = (1 - cos θ) / (1 + cos θ) = RHS.

Prove the identity: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

LHS = (sin² A + cosec² A + 2 sin A cosec A) + (cos² A + sec² A + 2 cos A sec A). Since sin A cosec A = 1 and cos A sec A = 1, LHS = (sin² A + cos² A) + (1 + cot² A) + (1 + tan² A) + 2 + 2 = 1 + 1 + cot² A + 1 + tan² A + 4 = 7 + tan² A + cot² A = RHS.

NCERT Solutions Class 10 Maths Chapter 8: Introduction to Trigonometry

Q: Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Using trigonometric identities: (1) sin A = 1/√(1 + cot² A), (2) tan A = 1/cot A, and (3) sec A = √(cot² A + 1)/cot A.

Detailed solutions for Exercise 8.3 (Trigonometric Identities).

Ex 8.1 Ex 8.2 Ex 8.3

#### **Q1. Express the trigonometric ratios $\sin A, \sec A$ and $\tan A$ in terms of $\cot A$.** #### **A1. Solution:** * **For $\sin A$:** We know that $\text{cosec}^2 A = 1 + \cot^2 A$ $\frac{1}{\sin^2 A} = 1 + \cot^2 A$ $\sin^2 A = \frac{1}{1 + \cot^2 A}$ **$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$** * **For $\tan A$:** **$\tan A = \frac{1}{\cot A}$** * **For $\sec A$:** $\sec^2 A = 1 + \tan^2 A$ $\sec^2 A = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A}$ **$\sec A = \frac{\sqrt{\cot^2 A + 1}}{\cot A}$** #### **Q2. Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.** #### **A2. Solution:** * **$\cos A = \frac{1}{\sec A}$** * **$\sin A$:** $\sin^2 A = 1 - \cos^2 A = 1 - \frac{1}{\sec^2 A} = \frac{\sec^2 A - 1}{\sec^2 A} \Rightarrow \mathbf{\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}}$ * **$\tan A$:** $1 + \tan^2 A = \sec^2 A \Rightarrow \mathbf{\tan A = \sqrt{\sec^2 A - 1}}$ * **$\cot A$:** $\cot A = \frac{1}{\tan A} \Rightarrow \mathbf{\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}}$ * **$\text{cosec } A$:** $\text{cosec } A = \frac{1}{\sin A} \Rightarrow \mathbf{\text{cosec } A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}}$ #### **Q3. Choose the correct option. Justify your choice.** (i) $9 \sec^2 A - 9 \tan^2 A =$ (ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta - \text{cosec } \theta) =$ (iii) $(\sec A + \tan A) (1 - \sin A) =$ (iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$ #### **A3. Solution:** (i) **Solution:** $9(\sec^2 A - \tan^2 A) = 9(1) = 9$. **(Option B)** (ii) **Solution:** Converting to sin and cos: $(1 + \frac{s}{c} + \frac{1}{c})(1 + \frac{c}{s} - \frac{1}{s}) = \frac{(c+s+1)(s+c-1)}{sc} = \frac{(s+c)^2 - 1^2}{sc} = \frac{s^2+c^2+2sc-1}{sc} = \frac{1+2sc-1}{sc} = 2$. **(Option C)** (iii) **Solution:** $(\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A) = \frac{(1+\sin A)(1-\sin A)}{\cos A} = \frac{1-\sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A$. **(Option D)** (iv) **Solution:** $\frac{\sec^2 A}{\text{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$. **(Option D)** #### **Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.** (i) $(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$ (ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$ (iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$ (vi) $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$ (viii) $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$ *(Note: Parts iv, v, vii, ix, x are also part of this question in the textbook but omitted here for brevity unless requested).* #### **A4. Solution:** (i) **Solution:** LHS $= (\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta})^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$ $= \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta} = \text{RHS.}$ (ii) **Solution:** LHS $= \frac{\cos^2 A + (1 + \sin A)^2}{\cos A(1 + \sin A)} = \frac{\cos^2 A + 1 + \sin^2 A + 2\sin A}{\cos A(1 + \sin A)}$ $= \frac{(1) + 1 + 2\sin A}{\cos A(1 + \sin A)} = \frac{2 + 2\sin A}{\cos A(1 + \sin A)} = \frac{2(1 + \sin A)}{\cos A(1 + \sin A)} = \frac{2}{\cos A} = 2\sec A = \text{RHS.}$ (iii) **Solution:** Convert to sin/cos: $\frac{s/c}{1 - c/s} + \frac{c/s}{1 - s/c} = \frac{s^2}{c(s-c)} + \frac{c^2}{s(c-s)} = \frac{s^2}{c(s-c)} - \frac{c^2}{s(s-c)}$ $= \frac{s^3 - c^3}{sc(s-c)} = \frac{(s-c)(s^2+sc+c^2)}{sc(s-c)} = \frac{1+sc}{sc} = \frac{1}{sc} + 1 = \sec \theta \text{cosec } \theta + 1 = \text{RHS.}$ (vi) **Solution:** LHS $= \sqrt{\frac{1 + \sin A}{1 - \sin A} \times \frac{1 + \sin A}{1 + \sin A}} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$ $= \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS.}$ (viii) **Solution:** LHS $= (\sin^2 A + \text{cosec}^2 A + 2\sin A \text{cosec } A) + (\cos^2 A + \sec^2 A + 2\cos A \sec A)$ $= (\sin^2 A + \cos^2 A) + (1 + \cot^2 A) + 2(1) + (1 + \tan^2 A) + 2(1)$ $= 1 + 1 + \cot^2 A + 2 + 1 + \tan^2 A + 2 = 7 + \tan^2 A + \cot^2 A = \text{RHS.}$