UPSC NDA 2 2021 Mathematics Solved Paper

Concept:Use inverse trigonometric identities to simplify the given expression into a form where the $\cot$ and $\cot^{-1}$ cancel out.Explanation:Let $E = \cot\left[ \sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2} \right]$.First, rewrite $\cot^{-1} \frac{3}{2}$ as $\sin^{-1} \frac{2}{\sqrt{13}}$.For $\cot a = \frac{3}{2}$, adjacent side $= 3$, opposite side $= 2$, so hypotenuse $= \sqrt{13}$.Thus $a = \sin^{-1} \frac{2}{\sqrt{13}}$.Substitute into $E$: $E = \cot\left[ \sin^{-1} \frac{3}{5} + \sin^{-1} \frac{2}{\sqrt{13}} \right]$.Use the identity $\sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})$.Here $x = \frac{3}{5}$, $y = \frac{2}{\sqrt{13}}$.Compute $\sqrt{1-y^2} = \sqrt{1 - \frac{4}{13}} = \frac{3}{\sqrt{13}}$ and $\sqrt{1-x^2} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$.Argument sum: $\frac{3}{5} \cdot \frac{3}{\sqrt{13}} + \frac{2}{\sqrt{13}} \cdot \frac{4}{5} = \frac{9}{5\sqrt{13}} + \frac{8}{5\sqrt{13}} = \frac{17}{5\sqrt{13}}$.Hence, $E = \cot\left[ \sin^{-1} \frac{17}{5\sqrt{13}} \right]$.Let $\sin\theta = \frac{17}{5\sqrt{13}}$. Opposite side $= 17$, hypotenuse $= 5\sqrt{13}$.Adjacent side $= \sqrt{(5\sqrt{13})^2 - 17^2} = \sqrt{325 - 289} = 6$.Therefore, $\cot\theta = \frac{6}{17}$, so $\sin^{-1} \frac{17}{5\sqrt{13}} = \cot^{-1} \frac{6}{17}$.Finally, $E = \cot\left[ \cot^{-1} \frac{6}{17} \right] = \frac{6}{17}$.Answer:$\frac{6}{17}$