Question

The value of $\underset{k=1}{\overset{13}{∑}}\frac{1}{\sin \left(\frac{π}{4}+\frac{(k−1)π}{6}\right)\sin \left(\frac{π}{4}+\frac{kπ}{6}\right)}$ is equal to

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

Answer 1

Answer: (C)

$\underset{k=1}{\overset{13}{∑}}\frac{1}{\sin \left(\frac{π}{4}+(k−1)\frac{π}{6}\right)\sin \left(\frac{π}{4}+\frac{kπ}{6}\right)}$
$=2\underset{k=1}{\overset{13}{∑}}\frac{\sin \left[\left(\frac{π}{4}+\frac{kπ}{6}\right)−\left(\frac{π}{4}+(k−1)\frac{π}{6}\right)\right]}{\sin \left(\frac{π}{4}+(k−1)\frac{π}{6}\right)\sin \left(\frac{π}{4}+\frac{kπ}{6}\right)}$
$=2\underset{k=1}{\overset{13}{∑}}\left[\cot \left(\frac{π}{4}+(k−1)\frac{π}{6}\right)−\cot \left(\frac{π}{4}+\frac{kπ}{6}\right)\right]$
$=2[\cot \frac{π}{4}−\cot \left(\frac{π}{4}+\frac{π}{6}\right)+\cot \left(\frac{π}{4}+\frac{π}{6}\right)−\cot \left(\frac{π}{4}+\frac{2π}{6}\right)$
$+…+\cot \left(\frac{π}{4}+\frac{12π}{6}\right)−\cot \left(\frac{π}{4}+\frac{13π}{6}\right)]$
$=2\left[\cot \frac{π}{4}−\cot \left(\frac{π}{4}+\frac{13π}{6}\right)\right]$
$=2[1−(2−\sqrt{3})]$
$=2(\sqrt{3}−1)$