The value of sum_{k=1}^{13} frac{1}{sin left( | vedclass

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(C) Let $S = \sum_{k=1}^{13} \frac{1}{\sin \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$.Multiply a

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

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(C) Let $S = \sum_{k=1}^{13} \frac{1}{\sin \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$.Multiply and divide by $\sin(\frac{\pi}{6})$:$S = \frac{1}{\sin(\frac{\pi}{6})} \sum_{k=1}^{13} \frac{\sin(\frac{\pi}{6})}{\sin \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$.Since $\frac{\pi}{6} = (\frac{\pi}{4}+\frac{k \pi}{6}) - (\frac{\pi}{4}+(k-1) \frac{\pi}{6})$,we use the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:$S = 2 \sum_{k=1}^{13} \left[ \cot \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right) - \cot \left(\frac{\pi}{4}+\frac{k \pi}{6}\right) \right]$.This is a telescoping sum:$S = 2 \left[ \cot \frac{\pi}{4} - \cot \left(\frac{\pi}{4} + \frac{13\pi}{6}\right) \right]$.Since $\cot \frac{\pi}{4} = 1$ and $\cot(\frac{\pi}{4} + \frac{13\pi}{6}) = \cot(\frac{\pi}{4} + \frac{\pi}{6}) = \cot(\frac{5\pi}{12}) = 2-\sqrt{3}$:$S = 2 [1 - (2-\sqrt{3})] = 2(\sqrt{3}-1)$.

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

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