Question

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

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

Answer 1

Answer: (C)

Let $\alpha=\theta+(m-1) \frac{\pi}{6}$
$\& \beta=\theta+m \frac{\pi}{6}$
So, $\beta-\alpha=\frac{\pi}{6}$
Here, $\displaystyle\sum_{ m =1}^9 \sec \alpha \cdot \sec \beta=\sum_{ m =1}^9 \frac{1}{\cos \alpha \cdot \cos \beta}$
$=2 \displaystyle\sum_{ m =1}^9 \frac{\sin (\beta-\alpha)}{\cos \alpha \cdot \cos \beta}=2 \displaystyle\sum_{ m =1}^9(\tan \beta-\tan \alpha)$
$=2 \displaystyle\sum_{ m =1}^9\left(\tan \left(\theta+ m \frac{\pi}{6}\right)-\tan \left(\theta+( m -1) \frac{\pi}{6}\right)\right)$
$=2\left(\tan \left(\theta+\frac{9 \pi}{6}\right)-\tan \theta\right)=2(-\cot \theta-\tan \theta)=-\frac{8}{\sqrt{3}}$
(Given)
$\therefore \tan \theta+\cot \theta=\frac{4}{\sqrt{3}} $
$ \Rightarrow \tan \theta=\frac{1}{\sqrt{3}} \text { or } \sqrt{3}$
So, $S=\left\{\frac{\pi}{6}, \frac{\pi}{3}\right\}$
$\displaystyle\sum_{\theta \in S } \theta=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$