UPSC NDA I 2017 Mathematics Solved Paper

Concept:Use basic trigonometric identities and algebraic manipulation to relate $\sec\theta - \csc\theta$ to $\sin\theta - \cos\theta$.Explanation:Given: $\sec\theta - \csc\theta = \frac{4}{3}$.Rewrite in terms of $\sin\theta$ and $\cos\theta$: $\frac{1}{\cos\theta} - \frac{1}{\sin\theta} = \frac{4}{3}$.Combine the left side: $\frac{\sin\theta - \cos\theta}{\sin\theta \cos\theta} = \frac{4}{3}$. $\quad$ (1)Square both sides of (1): $\frac{(\sin\theta - \cos\theta)^2}{(\sin\theta \cos\theta)^2} = \frac{16}{9}$.Expand numerator: $(\sin\theta - \cos\theta)^2 = \sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta = 1 - 2\sin\theta\cos\theta$.Let $x = \sin\theta \cos\theta$. Then $\frac{1 - 2x}{x^2} = \frac{16}{9}$.Cross multiply: $9(1 - 2x) = 16x^2 \Rightarrow 9 - 18x = 16x^2$.Rearrange: $16x^2 + 18x - 9 = 0$. Factor: $(8x - 3)(2x + 3) = 0$.So $x = \frac{3}{8}$ or $x = -\frac{3}{2}$.Since $\sin\theta$ and $\cos\theta$ are between $-1$ and $1$, their product must satisfy $-1 \le x \le 1$. Thus $x = -\frac{3}{2}$ is impossible. Hence $x = \frac{3}{8}$.From equation (1): $\sin\theta - \cos\theta = \frac{4}{3} \cdot \sin\theta\cos\theta = \frac{4}{3} \times \frac{3}{8} = \frac{1}{2}$.Only $\frac{1}{2}$ is valid; $-2$ is not obtained.Answer:The correct value is $\frac{1}{2}$ only, which corresponds to option B.