The value of ∑k=1261sin(π4+(k−1)π6)sin(π4+kπ6) is equal to

∑k=1131sin(π4+(k−1)π6)sin(π4+kπ6) =∑k=113sin{(π4+kπ6)−(π4+(k−1)π6)}sin(π4+(k−1)π6)sin(π4+kπ6)·1sinπ6 =2∑k=113[cot(π4+(k−1)π6)−cot(π4+kπ6)] =2[cotπ4−cot(π4+13π3)]=2[cotπ4−cot7π12]=2[1−(3−2)]

The value of ∑k=1261sin(π4+(k−1)π6)sin(π4+kπ6) is equal to

∑k=1131sin(π4+(k−1)π6)sin(π4+kπ6) =∑k=113sin{(π4+kπ6)−(π4+(k−1)π6)}sin(π4+(k−1)π6)sin(π4+kπ6)·1sinπ6 =2∑k=113[cot(π4+(k−1)π6)−cot(π4+kπ6)] =2[cotπ4−cot(π4+13π3)]=2[cotπ4−cot7π12]=2[1−(3−2)]

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The value of $\sum_{k = 1}^{26} \frac{1}{\sin (\frac{π}{4} + \frac{(k - 1) π}{6}) \sin (\frac{π}{4} + \frac{k π}{6})}$ is equal to

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

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

$3 - \sqrt{3}$

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

answer is C.

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Detailed Solution

$\sum_{k = 1}^{13} \frac{1}{\sin (\frac{π}{4} + \frac{(k - 1) π}{6}) \sin (\frac{π}{4} + \frac{k π}{6})}$
$= \sum_{k = 1}^{13} \frac{\sin {(\frac{π}{4} + \frac{k π}{6}) - (\frac{π}{4} + \frac{(k - 1) π}{6})}}{\sin (\frac{π}{4} + \frac{(k - 1) π}{6}) \sin (\frac{π}{4} + \frac{k π}{6})} \cdot \frac{1}{\sin \frac{π}{6}}$
$= 2 \sum_{k = 1}^{13} [\cot (\frac{π}{4} + (k - 1) \frac{π}{6}) - \cot (\frac{π}{4} + \frac{k π}{6})]$
$= 2 [\cot \frac{π}{4} - \cot (\frac{π}{4} + \frac{13 π}{3})] = 2 [\cot \frac{π}{4} - \cot \frac{7 π}{12}] = 2 [1 - (\sqrt{3} - 2)]$