Mathematics - Sequence & Series Question with Solution | TestHub

Let$\frac{\pi }{4}+\left(K-1\right)\frac{\pi }{6}=\theta$
$\Rightarrow {\displaystyle \sum _{k=1}^{13}}\frac{1}{\sin \theta \sin \left(\mathrm{\theta }+\frac{\pi }{6}\right)}=\frac{1}{\sin \frac{\pi }{6}}{\displaystyle \sum _{k=1}^{13}}\frac{\sin \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\sin \theta \sin \left(\theta +\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)}$
$=2{\displaystyle \sum _{k=1}^{13}}\frac{\sin \left(\theta +\frac{\pi }{6}-\theta \right)}{\sin \theta \sin \left(\theta +\frac{\pi }{6}\right)}=2{\displaystyle \sum _{k=1}^{13}}\frac{\sin \left(\theta +\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)\cos \theta -\cos \left(\theta +\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)\sin \theta }{\sin \theta \sin \left(\theta +\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)}$
$=2{\displaystyle \sum _{k=1}^{13}}\left(\cot \theta -\cot \left(\theta +\frac{\pi }{6}\right)\right)=2{\displaystyle \sum _{k=1}^{13}}\left(\cot \left(\frac{\pi }{4}+\left(k-1\right)\frac{\pi }{6}\right)-\cot \left(\frac{\pi }{4}+\frac{k\pi }{6}\right)\right)$
$=2\left(\cot \frac{\pi }{4}-\cot \left(\frac{\pi }{4}+\frac{\pi }{6}\right)+\cot \left(\frac{\pi }{4}+\frac{\pi }{6}\right)-\cot \left(\frac{\pi }{4}+\frac{2\pi }{6}\right)+-----\cot \left(\frac{\pi }{4}+\frac{12\pi }{6}\right)-\cot \left(\frac{\pi }{4}+\frac{13\pi }{6}\right)\right)$
$=2\left[\cot \frac{\pi }{4}-\cot \left(\frac{\pi }{4}+\frac{13\pi }{6}\right)\right]=2\left[1-\cot \frac{29\pi }{12}\right]=2\left[1-\cot {75}^o\right]=2\left[1-\left(2-\sqrt{3}\right)\right]=2\left(\sqrt{3-1}\right)$