JEE Main 2025 (Online) 28th January Evening ShiftQuestion 1
If $\displaystyle\sum_{r=1}^{13}\left\{\dfrac{1}{\sin\left(\dfrac{\pi}{4}+(r-1)\dfrac{\pi}{6}\right)\sin\left(\dfrac{\pi}{4}+\dfrac{r\pi}{6}\right)}\right\}=a\sqrt{3}+b$, $a,b\in\mathbb{Z}$, then $a^{2}+b^{2}$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. The angles differ by $\frac\pi6$; each term is a difference of cotangents divided by $\sin\frac\pi6$. Formulas used. (1) $\cot A-\cot B=\dfrac{\sin(B-A)}{\sin A\sin B}$Step 1 - One term. With $B=A+\frac\pi6$, by (1): $$\frac{1}{\sin A\sin\left(A+\frac\pi6\right)}=\frac{\cot A-\cot\left(A+\frac\pi6\right)}{\sin\frac\pi6}=2\left[\cot A-\cot\left(A+\tfrac\pi6\right)\right].$$ Step 2 - Telescope. The angles run from $\frac\pi4$ to $\frac\pi4+\frac{13\pi}6$: $$\text{sum}=2\left[\cot\frac\pi4-\cot\left(\frac\pi4+\frac{13\pi}{6}\right)\right]=2\left[1-\cot\frac{5\pi}{12}\right].$$ Step 3 - $\cot75^\circ$. $\cot75^\circ=\tan15^\circ=2-\sqrt3$, so the sum is $2(\sqrt3-1)=2\sqrt3-2$. Step 4 - Value. $a=2$, $b=-2$, $a^2+b^2=8$. Answer: (C)