Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

$\frac{1}{\sin \frac{\pi }{6}}{\sum }_{r=1}^{13}\frac{\sin \left[\left(\frac{\pi }{4}+\frac{r\pi }{6}\right)-\left(\frac{\pi }{4}\right)-(r-1)\frac{\pi }{6}\right]}{\sin \left(\frac{\pi }{4}+(r-1)\frac{\pi }{6}\right)\sin \left(\frac{\pi }{4}+\frac{r\pi }{6}\right)}$ $\begin{array}{rl}& \frac{1}{\sin \frac{\pi }{6}}\sum _{\mathrm{r}=1}^{13}\left(\cot \left(\frac{\pi }{4}+(\mathrm{r}-1)\frac{\pi }{6}\right)-\cot \left(\frac{\pi }{4}+\frac{\mathrm{r}\pi }{6}\right)\right)\\ & =2\sqrt{3}-2=\alpha \sqrt{3}+\mathrm{b}\end{array}$ So $a^2+b^2=8$