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

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

Assuming $\frac{\pi }{4}=\theta ,\alpha =\frac{\pi }{6}$
$=\sum _{k=1}^{13}\frac{1}{\sin (\theta +k\alpha )\sin (\theta +(k-1\left)\alpha \right)}=\frac{1}{\sin \alpha }\sum _{k=1}^{13}\frac{\sin \alpha }{\sin (\theta +k\alpha )\sin (\theta +(k-1\left)\alpha \right)}=\frac{1}{\sin \alpha }\sum _{k=1}^{13}\frac{\sin \left[\right(\theta +k\alpha )-(\theta +(k-1)\alpha \left)\right]}{\sin (\theta +k\alpha )\sin (\theta +(k-1\left)\alpha \right)}=\frac{1}{\sin \frac{\pi }{6}}\sum _{k=1}^{13}\cot (\theta +(k-1\left)\alpha \right)-\cot (\theta +k\alpha )=2\sum _{k=1}^{13}\cot (\theta +(k-1\left)\alpha \right)-\cot (\theta +k\alpha )=2[\cot \theta -\cot (\theta +13\alpha \left)\right]=2[\cot \frac{\pi }{4}-\cot (\frac{\pi }{4}+\frac{13\pi }{6})]=2[1-\cot \left(\frac{5\pi }{12}\right)]=2[1-\cot {75}^{\circ }]=2[1-(2-\sqrt{3}\left)\right]=2(\sqrt{3}-1)$