UPSC NDA I 2024 Mathematics Paper

Let's solve the given expression step by step:The expression is:$\frac{\sqrt{3} \cos 10^{\circ} - \sin 10^{\circ}}{\sin 25^{\circ} \cos 25^{\circ}}$First, we need to simplify the denominator. Recall that:$\sin 25^{\circ} \cos 25^{\circ} = \frac{1}{2} \sin 50^{\circ}$Thus, the expression becomes:$\frac{\sqrt{3} \cos 10^{\circ} - \sin 10^{\circ}}{\frac{1}{2} \sin 50^{\circ}} = 2 \frac{\sqrt{3} \cos 10^{\circ} - \sin 10^{\circ}}{\sin 50^{\circ}}$Next, let's consider the numerator. Note that:$\sin 50^{\circ} = \cos (90^{\circ} - 50^{\circ}) = \cos 40^{\circ}$Thus, we need to express the numerator in terms of angles close to 40 degrees. We also know that:$\cos 40^{\circ} = \cos (180^{\circ} - 140^{\circ}) = -\cos 140^{\circ}$Let's use the identity for the cosine of a sum:$\cos 40^{\circ} = \cos (10^{\circ} + 30^{\circ}) = \cos 10^{\circ} \cos 30^{\circ} - \sin 10^{\circ} \sin 30^{\circ}$We know that:$\cos 30^{\circ} = \frac{\sqrt{3}}{2} \quad \text{ and } \quad \sin 30^{\circ} = \frac{1}{2}$ So:$\cos 40^{\circ} = \cos 10^{\circ} \cdot \frac{\sqrt{3}}{2} - \sin 10^{\circ} \cdot \frac{1}{2} \Rightarrow 2 \cos 40^{\circ} = \sqrt{3} \cos 10^{\circ} - \sin 10^{\circ}$This means our numerator simplifies to:$2 \cos 40^{\circ}$Thus the entire expression becomes:$2 \cdot \frac{2 \cos 40^{\circ}}{\sin 50^{\circ}} = 2 \cdot 2 = 4$