UPSC CDS 1 2022 Math Paper

Concept:Use trigonometric identities and test a specific angle that satisfies the equation and lies in the given range.Explanation:Given: $\tan^2 \theta + 3 \sec \theta - 9 = 0$, with $0^\circ < \theta < 90^\circ$.We need $\tan^2 \theta + 3 \sec \theta$ to equal $9$.Since $9$ is rational, $\tan^2 \theta$ and $\sec \theta$ should both be rational for the sum to be rational.The only standard angle in $(0^\circ, 90^\circ)$ where $\sec \theta$ is rational is $\theta = 60^\circ$ (because $\sec 60^\circ = 2$).Check: $\tan 60^\circ = \sqrt{3}$, so $\tan^2 60^\circ = 3$.Then left side becomes $3 + 3 \times 2 - 9 = 3 + 6 - 9 = 0$. Hence $\theta = 60^\circ$ satisfies the equation.Now compute $12 \cot^2 \theta + 3 \csc \theta$ at $\theta = 60^\circ$:$\cot 60^\circ = \frac{1}{\sqrt{3}}$, so $\cot^2 60^\circ = \frac{1}{3}$.$\csc 60^\circ = \frac{2}{\sqrt{3}}$.Therefore, $12 \times \frac{1}{3} + 3 \times \frac{2}{\sqrt{3}} = 4 + \frac{6}{\sqrt{3}} = 4 + 2\sqrt{3}$.Rewrite $4 + 2\sqrt{3}$ as $(\sqrt{3})^{2} + (1)^{2} + 2(\sqrt{3})(1) = (\sqrt{3} + 1)^{2}$.Answer:$$(\sqrt{3} + 1)^{2}$$ which corresponds to option A.