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

$\sum _{k=1}^{13}\frac{1}{\sin (\frac{\pi }{4}+\frac{(k-1)\pi }{6})\sin (\frac{\pi }{4}+\frac{k\pi }{6})}$

$=2\sum _{k=1}^{13}\frac{\sin [(\frac{\pi }{4}+\frac{k\pi }{6})-(\frac{\pi }{4}+\frac{(k-1)\pi }{6})]}{\sin (\frac{\pi }{4}+\frac{(k-1)\pi }{6})\sin (\frac{\pi }{4}+\frac{k\pi }{6})}$

$=2\sum _{k=1}^{13}\frac{\sin (\frac{\pi }{4}+\frac{k\pi }{6}).\cos (\frac{\pi }{4}+\frac{(k-1)\pi }{6})-\sin (\frac{\pi }{4}+\frac{(k-1)\pi }{6})\cos (\frac{\pi }{4}+\frac{k\pi }{6})}{\sin (\frac{\pi }{4}+\frac{(k-1)\pi }{6})\sin (\frac{\pi }{4}+\frac{k\pi }{6})}$
$=2\sum _{k=1}^{13}[\cot (\frac{\pi }{4}+\frac{(k-1)\pi }{6})-\cot (\frac{\pi }{4}+\frac{k\pi }{6})]$

$=2[\cot \frac{\pi }{4}-\cot (\frac{\pi }{4}+\frac{13\pi }{6})]=2[\cot \frac{\pi }{4}-\cot (2\pi +\frac{\pi }{4}+\frac{\pi }{6})]$
$=2[1-\cot (\frac{\pi }{4}+\frac{\pi }{6})]=2[1-\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}]=2[1-\frac{\sqrt{3}-1}{\sqrt{3}+1}]=2\times \frac{2}{\sqrt{3}+1}=2(\sqrt{3}-1)$