If then is equal to : [2025]

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Solution

Q.
If $\sum_{r = 1}^{13} {\frac{1}{\sin (\frac{π}{4} + (r - 1) \frac{π}{6}) \sin (\frac{π}{4} + \frac{r π}{6})}} = a \sqrt{3} + b, b \in Z,$ then $a^{2} + b^{2}$ is equal to : [2025]

1 2

2 4

3 10

4 8

Ans.
(4)

The given expression can be written as,

$\frac{1}{\sin \frac{π}{6}} \sum_{r = 1}^{13} \frac{\sin [(\frac{π}{4} + \frac{r π}{6}) - [(\frac{π}{4}) + (r - 1) \frac{π}{6}]]}{\sin (\frac{π}{4} + (r - 1) \frac{π}{6}) \sin (\frac{π}{4} + \frac{r π}{6})}$

$= \frac{1}{\sin \frac{π}{6}} \sum_{r = 1}^{13} (\cot (\frac{π}{4} + (r - 1) \frac{π}{6}) - \cot (\frac{π}{4} + \frac{rπ}{6}))$

$[∵ \frac{\sin (A - B)}{\sin A \sin B} = \cot B - \cot A]$

$= 2 [\cot \frac{π}{4} - \cot (\frac{π}{4} + \frac{13 π}{6})] = 2 (1 - 2 + \sqrt{3})$

$= 2 \sqrt{3} - 2 = a \sqrt{3} + b$ [Given]

$∴ a = 2, b = - 2$

So, $a^{2} + b^{2} = 8$

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