Question
माना $S=\left\{\theta \in\left(0, \frac{\pi}{2}\right): \sum \limits_{m=1}^9 \sec \left(\theta+( m -1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{ m \pi}{6}\right)=-\frac{8}{\sqrt{3}}\right\}$ है। तब
$\beta=\theta+m \frac{\pi}{6}$
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, $S=\left\{\frac{\pi}{6}, \frac{\pi}{3}\right\}$
$\sum_{\theta \in S} \theta=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.