UPSC NDA 2 2017 Mathematics Solved Paper

Concept:Simplify $\sqrt{3}\csc 20^\circ - \sec 20^\circ$ by converting to sine and cosine, then using trigonometric identities.Explanation:Start with $\sqrt{3}\csc 20^\circ - \sec 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}$.Multiply numerator and denominator by $2$: $2\left(\frac{\frac{\sqrt{3}}{2}}{\sin 20^\circ} - \frac{\frac{1}{2}}{\cos 20^\circ}\right)$.Recognize $\frac{\sqrt{3}}{2} = \cos 30^\circ$ and $\frac{1}{2} = \sin 30^\circ$: $2\left(\frac{\cos 30^\circ}{\sin 20^\circ} - \frac{\sin 30^\circ}{\cos 20^\circ}\right)$.Combine over common denominator: $2\left(\frac{\cos 30^\circ \cos 20^\circ - \sin 30^\circ \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}\right)$.Use identity $\cos A \cos B - \sin A \sin B = \cos(A+B)$; numerator becomes $\cos(30^\circ+20^\circ) = \cos 50^\circ$.Denominator is $\sin 20^\circ \cos 20^\circ$. Multiply numerator and denominator by $2$ to apply double-angle: $2 \cdot 2 \cdot \frac{\cos 50^\circ}{2\sin 20^\circ \cos 20^\circ} = 4 \cdot \frac{\cos 50^\circ}{\sin 40^\circ}$ (since $2\sin 20^\circ \cos 20^\circ = \sin 40^\circ$).Now $\sin 40^\circ = \sin(90^\circ - 50^\circ) = \cos 50^\circ$ using $\sin(90^\circ-\theta)=\cos\theta$.Thus $\frac{\cos 50^\circ}{\sin 40^\circ} = \frac{\cos 50^\circ}{\cos 50^\circ} = 1$, so the expression equals $4 \times 1 = 4$.Answer:$4$ (Option A).