UPSC NDA I and II 2020 Math Solved Paper

Concept:Use trigonometric identities and the componendo-dividendo rule to express $\tan^{2} A$ in terms of $K$.Explanation:We are given $\frac{\tan 3A}{\tan A} = K$. Rewrite as $\frac{\sin 3A}{\cos 3A} \cdot \frac{\cos A}{\sin A} = K$, i.e., $\frac{\sin 3A \cos A}{\cos 3A \sin A} = \frac{K}{1}$. Apply componendo-dividendo: $\frac{\sin 3A \cos A - \cos 3A \sin A}{\sin 3A \cos A + \cos 3A \sin A} = \frac{K-1}{K+1}$. The numerator is $\sin(3A - A) = \sin 2A$, the denominator is $\sin(3A + A) = \sin 4A$. Thus $\frac{\sin 2A}{\sin 4A} = \frac{K-1}{K+1}$. Use $\sin 4A = 2\sin 2A \cos 2A$, so $\frac{\sin 2A}{2\sin 2A \cos 2A} = \frac{K-1}{K+1}$, giving $\frac{1}{2\cos 2A} = \frac{K-1}{K+1}$. Write $\cos 2A = \cos^{2} A - \sin^{2} A$. Then $\frac{1}{2(\cos^{2} A - \sin^{2} A)} = \frac{K-1}{K+1}$. Divide numerator and denominator by $\cos^{2} A$: $\frac{\sec^{2} A}{2(1 - \tan^{2} A)} = \frac{K-1}{K+1}$. Since $\sec^{2} A = 1 + \tan^{2} A$, we have $\frac{1 + \tan^{2} A}{1 - \tan^{2} A} = \frac{2K-2}{K+1}$. Apply componendo-dividendo again: $\frac{1 + \tan^{2} A + 1 - \tan^{2} A}{1 + \tan^{2} A - (1 - \tan^{2} A)} = \frac{(2K-2) + (K+1)}{(2K-2) - (K+1)}$. Simplify: $\frac{2}{2\tan^{2} A} = \frac{3K-1}{K-3}$, so $\frac{1}{\tan^{2} A} = \frac{3K-1}{K-3}$. Hence $\tan^{2} A = \frac{K-3}{3K-1}$.Answer:$\tan^{2} A = \frac{K-3}{3K-1}$ which corresponds to option B.