Let S= (0, 2 ): _ m=1 ^ 9 ( +(m-1) 6 ) ( +m 6 )=-8 3 Then. - Vidyadip

MCQ

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.

A
$S =\left\{\frac{\pi}{12}\right\}$
B
$S =\left\{\frac{2 \pi}{3}\right\}$
✓
$\sum_{\theta \in S} \theta=\frac{\pi}{2}$
D
$\sum_{\theta \in S} \theta=\frac{3 \pi}{4}$

✓

Answer

Correct option: C.

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

c
Let $\alpha=\theta+(m-1) \frac{\pi}{6}$

$\beta=\theta+m \frac{\pi}{6}$

So, $\beta-\alpha=\frac{\pi}{6}$

Here,$\sum_{m=1}^{9} \sec \alpha \cdot \sec \beta=\sum_{m=1}^{9} \frac{1}{\cos \alpha \cdot \cos \beta}$

$= 2 \sum_{m=1}^{9} \frac{\sin (\beta-\alpha)}{\cos \alpha \cdot \cos \beta}=2 \sum_{m=1}^{9}(\tan \beta-\tan \alpha)$

$= 2 \sum_{m=1}^{9}\left(\tan \left(\theta+m \frac{\pi}{6}\right)-\tan \left(\theta+(m-1) \frac{\pi}{6}\right)\right)$

$=2\left(\tan \left(\theta+\frac{9 \pi}{6}\right)-\tan \theta\right)=2(-\cot \theta-\tan \theta)=-\frac{8}{\sqrt{3}}$

(Given)

$\therefore \quad \tan \theta+\cot \theta=\frac{4}{\sqrt{3}}$

$\tan \theta=\frac{1}{\sqrt{3}}$ or $\sqrt{3}$

So, $S=\left\{\frac{\pi}{6}, \frac{\pi}{3}\right\}$

$\sum_{\theta \in S} \theta=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$

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