Let S = left{ theta in left( 0, frac{pi}{2} r | vedclass

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(C) Let $\alpha = 	heta + (m-1) \frac{\pi}{6}$ and $\beta = 	heta + m \frac{\pi}{6}$.Then $\beta - \alpha = \frac{\pi}{6}$.The sum is given by $\sum

Vedclass · Accepted Answer

$\sum_{	heta \in S} 	heta = \frac{\pi}{2}$

Vedclass · Answer

(C) Let $\alpha = 	heta + (m-1) \frac{\pi}{6}$ and $\beta = 	heta + m \frac{\pi}{6}$.Then $\beta - \alpha = \frac{\pi}{6}$.The sum is given by $\sum_{m=1}^{9} \sec \alpha \sec \beta = \sum_{m=1}^{9} \frac{1}{\cos \alpha \cos \beta}$.Multiplying and dividing by $\sin(\beta - \alpha) = \sin(\frac{\pi}{6}) = \frac{1}{2}$,we get:$2 \sum_{m=1}^{9} \frac{\sin(\beta - \alpha)}{\cos \alpha \cos \beta} = 2 \sum_{m=1}^{9} (	an \beta - 	an \alpha)$.This is a telescoping sum: $2 \left( 	an \left( 	heta + \frac{9\pi}{6} ight) - 	an 	heta ight) = 2 (	an(	heta + \frac{3\pi}{2}) - 	an 	heta) = 2(-\cot 	heta - 	an 	heta)$.Given $2(-\cot 	heta - 	an 	heta) = -\frac{8}{\sqrt{3}}$,we have $	an 	heta + \cot 	heta = \frac{4}{\sqrt{3}}$.Let $x = 	an 	heta$,then $x + \frac{1}{x} = \frac{4}{\sqrt{3}} \implies \sqrt{3}x^2 - 4x + \sqrt{3} = 0$.Solving the quadratic: $(\sqrt{3}x - 1)(x - \sqrt{3}) = 0$,so $	an 	heta = \frac{1}{\sqrt{3}}$ or $	an 	heta = \sqrt{3}$.Thus,$	heta = \frac{\pi}{6}$ or $	heta = \frac{\pi}{3}$.$S = \left\{ \frac{\pi}{6}, \frac{\pi}{3} ight\}$,so $\sum_{	heta \in S} 	heta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.

Let $S = \left\{ \theta \in \left( 0, \frac{\pi}{2} \right) : \sum_{m=1}^{9} \sec \left( \theta + (m-1) \frac{\pi}{6} \right) \sec \left( \theta + \frac{m \pi}{6} \right) = -\frac{8}{\sqrt{3}} \right\}$. Then:

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