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

The value of ∑ₖ₌₁¹³ 1/(sin(π/4 + ((k-1)π)/6)sin(π/4 + kπ/6)) is equal to

$\displaystyle\sum_{k=1}^{13} \dfrac{1}{\sin\left(\frac{\pi}{4}+\frac{(k-1)\pi}{6}\right)\sin\left(\frac{\pi}{4}+\frac{k\pi}{6}\right)}$

$\text{the two angles differ by } \frac{\pi}{6}$

$\text{To find: the value of the sum (choose the correct option)}$

$1.\ \text{write } \theta_k = \frac{\pi}{4}+\frac{(k-1)\pi}{6}$

$2.\ \text{use } \frac{1}{\sin x\sin(x+d)} = \frac{\cot x-\cot(x+d)}{\sin d}$

$3.\ \text{telescope and evaluate}$

$\theta_k = \frac{\pi}{4}+\frac{(k-1)\pi}{6},\qquad \theta_{k+1} = \theta_k + \frac{\pi}{6}$

$a_k = \frac{1}{\sin\theta_k\,\sin\theta_{k+1}}$

$\cot x - \cot(x+d) = \frac{\cos x}{\sin x} - \frac{\cos(x+d)}{\sin(x+d)}$

$\frac{\cos x\sin(x+d) - \cos(x+d)\sin x}{\sin x\sin(x+d)} = \frac{\sin d}{\sin x\sin(x+d)}$

$\therefore \frac{1}{\sin x\sin(x+d)} = \frac{\cot x - \cot(x+d)}{\sin d}$

$\sin d = \sin\frac{\pi}{6} = \frac{1}{2} \ \Rightarrow\ \frac{1}{\sin d} = 2$

$a_k = 2\left[\cot\theta_k - \cot\theta_{k+1}\right]$

$S = 2\displaystyle\sum_{k=1}^{13}\left[\cot\theta_k - \cot\theta_{k+1}\right]$

$S = 2\left[\cot\theta_1 - \cot\theta_{14}\right]$

$\theta_1 = \frac{\pi}{4},\qquad \theta_{14} = \frac{\pi}{4}+13\cdot\frac{\pi}{6}$

$\theta_{14} = \frac{\pi}{4}+\frac{13\pi}{6} = 2\pi + \frac{5\pi}{12}$

$\cot\theta_1 = \cot\frac{\pi}{4} = 1$

$\cot\theta_{14} = \cot\left(2\pi+\frac{5\pi}{12}\right) = \cot\frac{5\pi}{12}$

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

$\tan\frac{\pi}{12} = \tan\left(\frac{\pi}{4}-\frac{\pi}{6}\right) = \frac{\tan\frac{\pi}{4}-\tan\frac{\pi}{6}}{1+\tan\frac{\pi}{4}\tan\frac{\pi}{6}}$

$= \frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}} = \frac{\sqrt3-1}{\sqrt3+1} = 2-\sqrt3$

$\Rightarrow \cot\theta_{14} = 2-\sqrt3$

$S = 2\left[\cot\theta_1 - \cot\theta_{14}\right] = 2\left[1-(2-\sqrt3)\right]$

$S = 2(\sqrt3-1)$

$S = 2(\sqrt3-1)$

$= \text{option } (3)$

$\boxed{S = 2(\sqrt3-1)}$

$\boxed{\text{option } (3)}$