UPSC NDA II 2018 Mathematics Solved Paper

Concept:Use the identity $\tan(270^\circ + \theta) = -\cot\theta$ and solve the trigonometric equation by checking known values of $\tan A$ in the fourth quadrant.Explanation:Step 1: Given: $3(3 - \tan^2 A - \cot A)^2 = 1$Divide both sides by 3: $(3 - \tan^2 A - \cot A)^2 = \frac{1}{3}$Take square root: $3 - \tan^2 A - \cot A = \pm\frac{1}{\sqrt{3}}$Since $A$ is in the fourth quadrant, $\tan A$ is negative and $\cot A$ is also negative. Test candidate angles from the options.Step 2: For $A = 300^\circ$, compute $\tan 300^\circ = \tan(270^\circ + 30^\circ) = -\cot 30^\circ = -\sqrt{3}$.Then $\tan^2 A = (-\sqrt{3})^2 = 3$ and $\cot A = \frac{1}{\tan A} = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}$.Step 3: Substitute into the left-hand side of the equation:$3 - \tan^2 A - \cot A = 3 - 3 - \left(-\frac{1}{\sqrt{3}}\right) = 0 + \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.Square it: $\left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$.Multiply by 3: $3 \times \frac{1}{3} = 1$, which matches the right-hand side.Step 4: Check other options quickly:• $315^\circ$ gives $\tan = -1$, then $3 - (-1)^2 - (-1) = 3 - 1 + 1 = 3$; squared = 9; times 3 = 27 ≠ 1.• $330^\circ$ gives $\tan = -\frac{1}{\sqrt{3}}$, then $3 - \frac{1}{3} - (-\sqrt{3}) = 3 - \frac{1}{3} + \sqrt{3} \neq \pm\frac{1}{\sqrt{3}}$.• $345^\circ$ gives $\tan \approx -0.2679$, not a standard value; similarly fails.Thus only $A = 300^\circ$ satisfies the equation.Answer:Option A: $300^\circ$