Let S={theta inleft(0, frac{pi}{2}right): sum | vedclass

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Question

Let $\alpha=	heta+(m-1) \frac{\pi}{6}$

$\beta=	heta+m \frac{\pi}{6}$

So, $\beta-\alpha=\frac{\pi}{6}$

Here,$\sum_{m=1}^{9} \sec \alpha \cdo

Vedclass · Accepted Answer

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

Vedclass · Answer

Let $\alpha=	heta+(m-1) \frac{\pi}{6}$

$\beta=	heta+m \frac{\pi}{6}$

So, $\beta-\alpha=\frac{\pi}{6}$

Here,$\sum_{m=1}^{9} \sec \alpha \cdot \sec \beta=\sum_{m=1}^{9} \frac{1}{\cos \alpha \cdot \cos \beta}$

$= 2 \sum_{m=1}^{9} \frac{\sin (\beta-\alpha)}{\cos \alpha \cdot \cos \beta}=2 \sum_{m=1}^{9}(	an \beta-	an \alpha)$

$= 2 \sum_{m=1}^{9}\left(	an \left(	heta+m \frac{\pi}{6}ight)-	an \left(	heta+(m-1) \frac{\pi}{6}ight)ight)$

$=2\left(	an \left(	heta+\frac{9 \pi}{6}ight)-	an 	hetaight)=2(-\cot 	heta-	an 	heta)=-\frac{8}{\sqrt{3}}$

(Given)

$	herefore \quad 	an 	heta+\cot 	heta=\frac{4}{\sqrt{3}}$

$	an 	heta=\frac{1}{\sqrt{3}}$ or $\sqrt{3}$

So, $S=\left\{\frac{\pi}{6}, \frac{\pi}{3}ight\}$

$\sum_{	heta \in S} 	heta=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$

Let $S={\theta \in\left(0, \frac{\pi}{2}\right): \sum_{m=1}^{9}}$

$\sec \left(\theta+(m-1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{m \pi}{6}\right)=-\frac{8}{\sqrt{3}}$ Then.

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