JEE Advanced 2025 · Paper 2, Mathematics question 15

Each term telescopes in part.
$$\frac{1}{\sin a\sin(a+1^\circ)}=\frac{\cot a-\cot(a+1^\circ)}{\sin1^\circ}.$$
So
$$\alpha\sin1^\circ=\sum_{\text{even }a=60}^{118}\cot a-\sum_{\text{odd }b=61}^{119}\cot b.$$

Use $\cot(180^\circ-x)=-\cot x$.

• In the odd sum, the terms pair as 61↔119, …, 89↔91 and all cancel.
• In the even sum, the terms pair as 62↔118, …, 88↔92 and cancel; $\cot90^\circ=0$; only $\cot60^\circ=\frac1{\sqrt3}$ is left.

Answer. $\alpha=\frac{\operatorname{cosec}1^\circ}{\sqrt3}$, so
$$\left(\frac{\operatorname{cosec}1^\circ}{\alpha}\right)^2=3.$$