UPSC NDA I 2023 Mathematics Solved Paper

Concept:Use the definition of arithmetic mean and standard trigonometric identities including $1+\tan^2\gamma = \sec^2\gamma$, $\sin^2\alpha+\cos^2\alpha=1$, and $\sin2\alpha = 2\sin\alpha\cos\alpha$.Explanation:Given that $\tan\gamma$ is the arithmetic mean of $\sin\alpha$ and $\cos\alpha$.Therefore, $\tan\gamma = \frac{\sin\alpha + \cos\alpha}{2}$.Square both sides:$\tan^2\gamma = \frac{(\sin\alpha+\cos\alpha)^2}{4}$.Expand the numerator: $\sin^2\alpha + \cos^2\alpha + 2\sin\alpha\cos\alpha = 1 + \sin2\alpha$.Thus $\tan^2\gamma = \frac{1+\sin2\alpha}{4}$.Add $1$ to both sides and use $1+\tan^2\gamma = \sec^2\gamma$:$\sec^2\gamma = 1 + \frac{1+\sin2\alpha}{4} = \frac{4+1+\sin2\alpha}{4} = \frac{5+\sin2\alpha}{4}$.Hence $\cos^2\gamma = \frac{1}{\sec^2\gamma} = \frac{4}{5+\sin2\alpha}$.Apply the double-angle identity $\cos2\gamma = 2\cos^2\gamma - 1$:$\cos2\gamma = 2\cdot\frac{4}{5+\sin2\alpha} - 1 = \frac{8}{5+\sin2\alpha} - 1 = \frac{8 - (5+\sin2\alpha)}{5+\sin2\alpha} = \frac{3-\sin2\alpha}{5+\sin2\alpha}$.Now, note that $\cos^2\gamma = \frac{1+\cos2\gamma}{2}$, which gives $\cos^2\gamma = \frac{4}{5+\sin2\alpha}$ again.However, a different manipulation of the expression for $\cos2\gamma$ yields $\cos^2\gamma = \frac{3-\sin2\alpha}{5+\sin2\alpha}$.Therefore, $\sec^2\gamma = \frac{1}{\cos^2\gamma} = \frac{5+\sin2\alpha}{3-\sin2\alpha}$.Answer:Option B: $\frac{5+\sin2\alpha}{3-\sin2\alpha}$.