The value of ∑k=1131sin(π4+(k−1)π6).sin(π4+kπ6) is equal to

1sinA−sinB=sin(A−B)sin(A−B).sinA.sinB =1sin(A−B)×sin(A−B)sinA.sinB=1sin(A−B)×[cotB−cotA] =1sinπ6∑K=113[cot(π4+(K−1)π6)−cot(π4+Kπ6)] =2[cotπ4−cot(π4+13π6)]=2[1−(2−3)]=2(3−1)

The value of ∑k=1131sin(π4+(k−1)π6).sin(π4+kπ6) is equal to

1sinA−sinB=sin(A−B)sin(A−B).sinA.sinB =1sin(A−B)×sin(A−B)sinA.sinB=1sin(A−B)×[cotB−cotA] =1sinπ6∑K=113[cot(π4+(K−1)π6)−cot(π4+Kπ6)] =2[cotπ4−cot(π4+13π6)]=2[1−(2−3)]=2(3−1)

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

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$3 - \sqrt{3}$

$2 (3 - \sqrt{3})$

$2 (\sqrt{3} - 1)$

$2 (2 + \sqrt{3})$

answer is C.

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Detailed Solution

$\frac{1}{\sin A - \sin B} = \frac{\sin (A - B)}{\sin (A - B) . \sin A . \sin B} \begin{array}{l} = \frac{1}{\sin (A - B)} \times \frac{\sin (A - B)}{\sin A . \sin B} \\ = \frac{1}{\sin (A - B)} \times [\cot B - \cot A] \end{array} = \frac{1}{\sin \frac{π}{6}} \sum_{K = 1}^{13} [\cot (\frac{π}{4} + \frac{(K - 1) π}{6}) - \cot (\frac{π}{4} + \frac{K π}{6})] \begin{array}{l} = 2 [\cot \frac{π}{4} - \cot (\frac{π}{4} + \frac{13 π}{6})] \\ = 2 [1 - (2 - \sqrt{3})] = 2 (\sqrt{3} - 1) \end{array}$