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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)
$(A)$ $\frac{\pi}{4}$ $(B)$ $\frac{\pi}{6}$ $(C)$ $\frac{\pi}{12}$ $(D)$ $\frac{5 \pi}{12}$
$(B,D)$
$(C,D)$
$(A,D)$
$(A,B)$
Solution
For $0 < \theta < \frac{\pi}{2}$
$\sum_{m=1}^6 \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}$
$\Rightarrow \sum_{m=1}^6 \frac{1}{\sin \left(\theta+\frac{(m-1) \pi}{4}\right) \sin \left(\theta+\frac{m \pi}{4}\right)}=4 \sqrt{2}$
$\Rightarrow \sum_{m=1}^6 \frac{\sin \left[\theta+\frac{m \pi}{4}-\left(\theta+\frac{(m-1) \pi}{4}\right)\right]}{\sin \frac{\pi}{4}\left\{\sin \left(\theta+\frac{(m-1) \pi}{4}\right) \sin \left(\theta+\frac{m \pi}{4}\right)\right\}}=4 \sqrt{2}$
$\Rightarrow \sum_{m=1}^6 \frac{\cot \left(\theta+\frac{(m-1) \pi}{4}\right)-\cot \left(\theta+\frac{m \pi}{4}\right)}{1 / \sqrt{2}}=4 \sqrt{2}$
$\Rightarrow \sum_{m=1}^6\left[\cot \left(\theta+\frac{(m-1) \pi}{4}\right)-\cot \left(\theta+\frac{m \pi}{4}\right)\right]=4$
$\Rightarrow \cot (\theta)-\cot \left(\theta+\frac{\pi}{4}\right)+\cot \left(\theta+\frac{\pi}{4}\right)-\cot \left(\theta+\frac{2 \pi}{4}\right)+$
$\cdots+\cot \left(\theta+\frac{5 \pi}{4}\right)-\cot \left(\theta+\frac{6 \pi}{4}\right)=4$
$\Rightarrow $$ \cot \theta-\cot \left(\frac{3 \pi}{2}+\theta\right)=4$
$\Rightarrow $$ \cot \theta+\tan \theta=4 \quad \Rightarrow \quad \tan ^2 \theta-4 \tan \theta+1=0$
$\Rightarrow $$ (\tan \theta-2)^2-3=0$
$\Rightarrow $$ (\tan \theta-2+\sqrt{3})(\tan \theta-2-\sqrt{3})=0$
$\Rightarrow $$ \tan \theta=2-\sqrt{3} \text { or } \tan \theta=2+\sqrt{3}$
$\Rightarrow $$ \theta=\frac{\pi}{12} ; \theta=\frac{5 \pi}{12}$
$\because $$ \theta \in\left(0, \frac{\pi}{2}\right) .$
