[JEE Advanced 2016] $\displaystyle\sum_{k=1}^{13}\frac1{\sin\!\left(\frac\pi4+\frac{(k-1)\pi}6\right)\sin\!\left(\frac\pi4+\frac{k\pi}6\right)}$ equals
$\displaystyle\sum_{k=1}^{13}\frac1{\sin\!\left(\frac\pi4+\frac{(k-1)\pi}6\right)\sin\!\left(\frac\pi4+\frac{k\pi}6\right)}$ equals
(a) $3-\sqrt3$
(b) $2(3-\sqrt3)$
(c) $2(\sqrt3-1)$
(d) $2(2-\sqrt3)$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Correct answer: (b) $2(3-\sqrt3)$
Using $\frac1{\sin A\sin(A+d)}=\frac{\cot A-\cot(A+d)}{\sin d}$ with $d=\frac\pi6$, the sum telescopes to $\frac{1}{\sin\frac\pi6}\big(\cot\frac\pi4-\cot(\frac\pi4+\frac{13\pi}6)\big)=2(3-\sqrt3)$.
JEE Advanced 2016 · Trigonometry — verified solution by the Vidaara Team.
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