The value of $\sum_{k=1}^{13} \frac{1}{\sin \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$ is equal to

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Sequences and Series

The value of

\sum_{k = 1}^{13} \frac{1}{s i n ( \frac{π}{4} + \frac{( k - 1 ) π}{6} ) s i n ( \frac{π}{4} + \frac{k π}{6} )}

is equal to

(A) $3 - 3$
(B) $2 (3 - 3)$
(C) $2 (3 - 1)$ ✓ Correct
(D) $2 (2 + 3)$

Answer: (C)

Solution

Use the identity

\frac{1}{s i n A s i n B} = \frac{1}{s i n ( B - A )} (cot A - cot B)

with

B - A = \frac{π}{6}

. The general term becomes

T_{k} = 2 (cot (\frac{π}{4} + (k - 1) \frac{π}{6}) - cot (\frac{π}{4} + k \frac{π}{6}))

. This is a telescoping series:

S_{13} = 2 (cot \frac{π}{4} - cot (\frac{π}{4} + \frac{13 π}{6})) = 2 (1 - cot \frac{5 π}{12})

. Since

cot \frac{5 π}{12} = 2 - 3

, we get

S_{13} = 2 (1 - (2 - 3)) = 2 (3 - 1)

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option

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