UPSC NDA I 2019 Mathematics Solved Paper

Concept:Use trigonometric identities for double angles to simplify the given expressions and compare coefficients.Explanation:Given $p = X \cos \theta - Y \sin \theta$ and $q = X \sin \theta + Y \cos \theta$.Compute $p^2 = X^2 \cos^2 \theta + Y^2 \sin^2 \theta - 2XY \cos \theta \sin \theta$.Compute $q^2 = X^2 \sin^2 \theta + Y^2 \cos^2 \theta + 2XY \cos \theta \sin \theta$.Compute $4pq = 4 \left( X^2 \cos \theta \sin \theta + XY \cos^2 \theta - XY \sin^2 \theta - Y^2 \sin \theta \cos \theta \right)$.Add the three expressions: $p^2 + 4pq + q^2 = X^2 ( \cos^2 \theta + \sin^2 \theta + 4 \cos \theta \sin \theta ) + Y^2 ( \sin^2 \theta + \cos^2 \theta - 4 \sin \theta \cos \theta ) + 4XY ( \cos^2 \theta - \sin^2 \theta )$.Using identities: $\cos^2 \theta + \sin^2 \theta = 1$, $2 \cos \theta \sin \theta = \sin 2\theta$, and $\cos^2 \theta - \sin^2 \theta = \cos 2\theta$.Thus $p^2 + 4pq + q^2 = X^2 (1 + 2 \sin 2\theta) + Y^2 (1 - 2 \sin 2\theta) + 4XY \cos 2\theta$.Given that this equals $AX^2 + BY^2$ (no $XY$ term), we must have $\cos 2\theta = 0$ and coefficients match: $A = 1 + 2 \sin 2\theta$, $B = 1 - 2 \sin 2\theta$.Since $\cos 2\theta = 0$, $2\theta = \frac{\pi}{2}$ so $\theta = \frac{\pi}{4}$. Then $\sin 2\theta = \sin \frac{\pi}{2} = 1$, giving $A = 3$, $B = -1$.Thus the required value of $\theta$ is $\frac{\pi}{4}$.Answer:$\theta = \frac{\pi}{4}$, which corresponds to option C.