If $\sum_{r=1}^{13}\left\{\frac{1}{\sin \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}\right) \sin \left(\f

Question

If $\sum_{r=1}^{13}\left\{\frac{1}{\sin \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}ight) \sin \left(\frac{\pi}{4}+\frac{r \pi}{6}ight)}ight\}$\ $=a \sqrt{3}+b, b \in \mathbf{Z}$, then $a^{2}+b^{2}$ is equal to:\

Professor PC Thomas Classes · Answer

(d): The given expression can be written as, $$ \begin{aligned} & \frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \frac{\sin \left[\left(\frac{\pi}{4}+\frac{r \pi}{6}ight)-\left[\left(\frac{\pi}{4}ight)+(r-1) \frac{\pi}{6}ight]ight]}{\sin \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}ight) \sin \left(\frac{\pi}{4}+\frac{r \pi}{6}ight)} \ & =\frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13}\left(\cot \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}ight)-\cot \left(\frac{\pi}{4}+\frac{r \pi}{6}ight)ight) \ & \quad\left[\because \frac{\sin (A-B)}{\sin A \sin B}=\cot B-\cot Aight] \ & =2\left[\cot \frac{\pi}{4}-\cot \left(\frac{\pi}{4}+\frac{13 \pi}{6}ight)ight]=2(1-2+\sqrt{3}) \ & =2 \sqrt{3}-2=a \sqrt{3}+b \ & 	herefore \quad a=2, b=-2 \end{aligned} $$ So, $a^{2}+b^{2}=8$\