If sum_{r=1}^{13} left{ frac{1}{sin left(frac | vedclass

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(C) Let $T_r = \frac{1}{\sin \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4} + \frac{r\pi}{6}\right)}$.Multiplying and divid

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$8$

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(C) Let $T_r = \frac{1}{\sin \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4} + \frac{r\pi}{6}\right)}$.Multiplying and dividing by $\sin \left(\frac{\pi}{6}\right)$:$T_r = \frac{1}{\sin \frac{\pi}{6}} \cdot \frac{\sin \left[ \left(\frac{\pi}{4} + \frac{r\pi}{6}\right) - \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) \right]}{\sin \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4} + \frac{r\pi}{6}\right)}$$T_r = 2 \left[ \cot \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) - \cot \left(\frac{\pi}{4} + \frac{r\pi}{6}\right) \right]$.Summing from $r=1$ to $13$ is a telescoping sum:$S = 2 \left[ \cot \left(\frac{\pi}{4}\right) - \cot \left(\frac{\pi}{4} + \frac{13\pi}{6}\right) \right]$.Since $\cot \left(\frac{\pi}{4}\right) = 1$ and $\cot \left(\frac{\pi}{4} + \frac{13\pi}{6}\right) = \cot \left(\frac{\pi}{4} + 2\pi + \frac{\pi}{6}\right) = \cot \left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \cot \left(\frac{5\pi}{12}\right) = 2 - \sqrt{3}$.$S = 2 [1 - (2 - \sqrt{3})] = 2 [-1 + \sqrt{3}] = 2\sqrt{3} - 2$.Comparing with $a\sqrt{3} + b$,we get $a = 2$ and $b = -2$.Thus,$a^2 + b^2 = 2^2 + (-2)^2 = 4 + 4 = 8$.

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

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