UPSC NDA I 2024 Mathematics Paper

Concept:Use the sine difference identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$ and the double-angle identity $\sin 2\theta = 2 \sin \theta \cos \theta$.Explanation:Start with $\frac{\sqrt{3} \cos 10^\circ - \sin 10^\circ}{\sin 25^\circ \cos 25^\circ}$.Multiply numerator and denominator by $\frac{1}{2}$.This gives $\frac{\frac{\sqrt{3}}{2} \cos 10^\circ - \frac{1}{2} \sin 10^\circ}{\frac{1}{2} \sin 25^\circ \cos 25^\circ}$.Recognise $\frac{\sqrt{3}}{2} = \sin 60^\circ$ and $\frac{1}{2} = \cos 60^\circ$.So numerator becomes $\sin 60^\circ \cos 10^\circ - \cos 60^\circ \sin 10^\circ = \sin(60^\circ - 10^\circ) = \sin 50^\circ$.Denominator becomes $\frac{1}{2} \sin 25^\circ \cos 25^\circ$.Hence the expression is $\frac{\sin 50^\circ}{\frac{1}{2} \sin 25^\circ \cos 25^\circ} = \frac{2 \sin 50^\circ}{\sin 25^\circ \cos 25^\circ}$.Now apply the double-angle identity: $\sin 50^\circ = \sin(2 \times 25^\circ) = 2 \sin 25^\circ \cos 25^\circ$.Substitute: numerator becomes $2 \times (2 \sin 25^\circ \cos 25^\circ) = 4 \sin 25^\circ \cos 25^\circ$.Cancel $\sin 25^\circ \cos 25^\circ$ in numerator and denominator.The result is $4$.Answer:Option D: $4$.