Question

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

• A. $3-\sqrt{3}$
• B. $2(3-\sqrt{3})$
• C. $2(\sqrt{3}-1)$
• D. $2(2-\sqrt{3})$

Answer 1

The correct option is C $2(\sqrt{3}-1)$

The value of
${\sum }_{k=1}^{13}\frac{1}{sin(\frac{\pi }{4}+(k-1\left)\frac{\pi }{6}\right)sin(\frac{\pi }{4}+\frac{k\pi }{6})}{\sum }_{k=1}^{13}\frac{2}{sin\frac{\pi }{6}}\left[\frac{sin\{\frac{\pi }{4}+\frac{k\pi }{6}-(\frac{\pi }{4}+(k-1\left)\frac{\pi }{6}\right)\}}{sin(\frac{\pi }{4}+(k-1\left)\frac{\pi }{6}\right))sin(\frac{\pi }{4}+\frac{k\pi }{6})}\right]{\sum }_{k=1}^{13}2[cot(\frac{\pi }{4}+(k-1\left)\frac{\pi }{6}\right)-cot(\frac{\pi }{4}+\frac{k\pi }{6})]=2[\{cot\frac{\pi }{4}-cot(\frac{\pi }{4}+\frac{\pi }{6})\}+\{cot(\frac{\pi }{4}+\frac{\pi }{6})-cot(\frac{\pi }{4}+\frac{2\pi }{6})\}+\dots +\{cot(\frac{\pi }{4}+\frac{12\pi }{6})-cot(\frac{\pi }{4}+\frac{13\pi }{6})\}]=2[cot\frac{\pi }{4}-cot(\frac{\pi }{4}+\frac{13\pi }{6})]=2[1-cot\frac{5\pi }{12}]=2[1-\frac{\sqrt{3}-1}{\sqrt{3}+1}]=2[1-(2-\sqrt{3}\left)\right]=2(\sqrt{3}-1)$