JEE Main 2025 Jan 28 Shift 2, Mathematics Q16: Summation of Series

Use the standard identity

$\frac{1}{\sin A\sin B}=\frac{1}{\sin (B-A)}(\cot A-\cot B)$,

with $A=\frac{\pi }{4}+(r-1)\frac{\pi }{6}$ and $B=A+\frac{\pi }{6}$, so $\sin (B-A)=\sin (\pi /6)=1/2$.

The sum telescopes:

${\sum }_{r=1}^{13}\frac{1}{\sin A_r\sin B_r}=\frac{1}{\sin (\pi /6)}[\cot (\pi /4)-\cot (\pi /4+13\pi /6)]=2[1-\cot (\pi /4+13\pi /6)]$.

Now $\frac{\pi }{4}+\frac{13\pi }{6}=\frac{29\pi }{12}$, and $\cot (29\pi /12)=\cot (29\pi /12-2\pi )=\cot (5\pi /12)=2-\sqrt{3}$.

Sum $=2[1-(2-\sqrt{3})]=2(\sqrt{3}-1)=2\sqrt{3}-2$.

Matching $a\sqrt{3}+b$: $a=2$, $b=-2$, so $a^2+b^2=4+4=8$.