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

$\sum _{k=1}^{13}\frac{\sin [(\frac{\pi }{4}+\frac{k\pi }{6})-(\frac{\pi }{4}+(k-1\left)\frac{\pi }{6}\right)]}{\sin \frac{\pi }{6}\left(\sin (\frac{\pi }{4}+\frac{k\pi }{6})\sin (\frac{\pi }{4}+(k-1\left)\frac{\pi }{6}\right)\right)}=2\sum _{k=1}^{13}(\cot (\frac{\pi }{4}+(k-1\left)\frac{\pi }{6}\right)-\cot (\frac{\pi }{4}+\frac{k\pi }{6}))$
using compound angle formula.
$=2(\cot \frac{\pi }{4}-\cot (\frac{\pi }{4}+\frac{13\pi }{6}))$

$=2(1-\cot \left(\frac{29\pi }{12}\right))$

$=2(1-\cot \left(\frac{5\pi }{12}\right))$

$=2(1-(2-\sqrt{3}\left)\right)$

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