UPSC NDA I 2017 Mathematics Solved Paper

Concept:Use the half‑angle identity $1 + \cos A = 2\cos^2\frac{A}{2}$ and determine the sign from the quadrant of $\frac{A}{2}$.Explanation:Given $\sin A = \frac{3}{5}$.$\cos A = \pm\sqrt{1-\sin^2 A} = \pm\sqrt{1-\frac{9}{25}} = \pm\frac{4}{5}$.Since $450^\circ < A < 540^\circ$, subtract $360^\circ$ to see that $A$ lies in the second quadrant.In the second quadrant, $\cos A$ is negative, so $\cos A = -\frac{4}{5}$.Apply the half‑angle formula: $1 + \cos A = 2\cos^2\frac{A}{2}$.$1 + \left(-\frac{4}{5}\right) = \frac{1}{5} = 2\cos^2\frac{A}{2}$.Thus $\cos^2\frac{A}{2} = \frac{1}{10}$ and $\cos\frac{A}{2} = \pm\frac{1}{\sqrt{10}}$.Now find the quadrant of $\frac{A}{2}$: $225^\circ < \frac{A}{2} < 270^\circ$, which is the third quadrant.In the third quadrant, $\cos$ is negative, so $\cos\frac{A}{2} = -\frac{1}{\sqrt{10}}$.This value is not listed among the given options.Answer:D. None of the above