UPSC CDS 2 2022 Math Paper

Concept:An identity is an equation that holds true for all allowed values of the variable. We simplify each expression to check if both sides are exactly equal.Explanation:Statement 1: Start with $\frac{\tan\theta + \sin\theta}{\tan\theta - \sin\theta}$.Write $\tan\theta = \frac{\sin\theta}{\cos\theta}$:$\frac{\frac{\sin\theta}{\cos\theta} + \sin\theta}{\frac{\sin\theta}{\cos\theta} - \sin\theta} = \frac{\sin\theta\left(\frac{1}{\cos\theta} + 1\right)}{\sin\theta\left(\frac{1}{\cos\theta} - 1\right)}$.Cancel $\sin\theta$ (since $0 < \theta < \frac{\pi}{2}$, $\sin\theta\neq0$):$\frac{\frac{1}{\cos\theta} + 1}{\frac{1}{\cos\theta} - 1} = \frac{\sec\theta + 1}{\sec\theta - 1}$.Thus, the left side equals the right side for all $\theta$ in the domain. Hence, Statement 1 is an identity.Statement 2: Start with $\frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin^2\theta}$.The denominator simplifies: $\cos^2\theta + \sin^2\theta = 1$. So the left side becomes $\cos^2\theta - \sin^2\theta$.Now simplify the right side: $\frac{2\tan\theta}{\tan^2\theta + 1}$.Use $\tan\theta = \frac{\sin\theta}{\cos\theta}$:$\frac{2\cdot\frac{\sin\theta}{\cos\theta}}{\frac{\sin^2\theta}{\cos^2\theta} + 1} = \frac{2\sin\theta/\cos\theta}{(\sin^2\theta + \cos^2\theta)/\cos^2\theta} = \frac{2\sin\theta}{\cos\theta} \cdot \frac{\cos^2\theta}{1} = 2\sin\theta\cos\theta = \sin 2\theta$.But the left side $\cos^2\theta - \sin^2\theta = \cos 2\theta$.Since $\sin 2\theta \neq \cos 2\theta$ in general (except specific angles), the two sides are not equal for all $\theta$. Therefore, Statement 2 is not an identity.Answer:Only Statement 1 is an identity. Hence, the correct option is A: Only 1.