For $0 < \theta < \frac{\pi}{2}$, the solution(s) of $\sum_{m=1}^6…
For $0 < \theta < \frac{\pi}{2}$, the solution(s) of $\sum_{m=1}^6 \operatorname{cosec}\left[\theta+\frac{(m-1) \pi}{4}\right]$ $\operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}$ is/are
- $\frac{\pi}{4}$
- $\frac{\pi}{6}$
- $\frac{\pi}{12}$
- $\frac{5 \pi}{12}$
Solution
For $0 < \theta < \frac{\pi}{2}$,
$
\begin{array}{r}
\sum_{m=1}^6 \operatorname{cosec}\left[\theta+\frac{(m-1) \pi}{4}\right] \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right) \\
=4 \sqrt{2} \\
\Rightarrow \sum_{m=1}^6 \frac{\sin \left[\theta+\frac{m \pi}{4}-\left(\theta+\frac{(m-1) \pi}{4}\right)\right]}{\sin \frac{\pi}{4}\left\{\sin \left(\theta+\frac{(m-1) \pi}{4}\right) \sin \left(\theta+\frac{m \pi}{4}\right)\right\}} \\
=4 \sqrt{2}
\end{array}
$
$
\begin{array}{r}
\Rightarrow \sum_{m=1}^6 \frac{\cot \left(\theta+\frac{(m-1) \pi}{4}\right)-\cot \left(\theta+\frac{m \pi}{4}\right)}{1 / 2} \\
=4 \sqrt{2}
\end{array}
$
$
\begin{aligned}
& \Rightarrow \quad \sum_{m=1}^6\left[\cot \left(\theta+\frac{(m-1) \pi}{4}\right)\right. \\
& \left.-\cot \left(\theta+\frac{m \pi}{4}\right)\right]=4 \\
& \Rightarrow \cot (\theta)-\cot \left(\theta+\frac{\pi}{4}\right)+\cot \left(\theta+\frac{\pi}{4}\right) \\
& -\cot \left(\theta+\frac{2 \pi}{4}\right)+\ldots+\cot \left(\theta+\frac{5 \pi}{4}\right) \\
& -\cot \left(\theta+\frac{6 \pi}{4}\right)=4 \\
& \Rightarrow \quad \cot \theta-\cot \left(\frac{3 \pi}{2}+\theta\right)=4 \\
& \Rightarrow \quad \cot \theta+\tan \theta=4 \\
& \Rightarrow \quad \tan ^2 \theta-4 \tan \theta+1=0 \\
& \Rightarrow \quad(\tan \theta-2)^2-3=0 \\
& \Rightarrow(\tan \theta-2+\sqrt{3})(\tan \theta-2-\sqrt{3})=0 \\
& \Rightarrow \quad \tan \theta=2-\sqrt{3} \text { or } \\
& \tan \theta=2+\sqrt{3} \\
& \Rightarrow \theta=\frac{\pi}{12} ; \theta=\frac{5 \pi}{12} \Rightarrow \theta \in\left(0, \frac{\pi}{2}\right) \\
&
\end{aligned}
$
Asked in: JEE Advanced 2009 (Paper 2)
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