Question

For $0 < \theta < \frac{\pi}{2}$, the solution(s) of $\displaystyle\sum _{m=1}^{6} cosec \Bigg(\theta+\frac{(m-1)\pi}{4}\Bigg)cosec\Bigg(\theta+\frac{m\pi}{4}\Bigg)=4\sqrt{2}$ is/are

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{6}$
• C. $\frac{\pi}{12}$
• D. $\frac{5\pi}{12}$

Answer 1

Answer: (D)

$\frac{5\pi}{12}$

Answer 2

For $0 < \theta < \frac{\pi}{2}$
$\displaystyle\sum _{m=1}^{6}cosec\Bigg(\theta+\frac{(m-1)}{4}\Bigg)cosec\Bigg(\theta+frac{m\pi}{4}\Bigg)=4\sqrt{2}$
\Rightarrow \hspace25mm \displaystyle\sum _{m=1}^{6}\frac{1}{sin\Big(\theta+\frac{(m-1)\pi}{4}\Big)sin\Big(\theta+\frac{m\pi}{4}\Big)}=4\sqrt{2}
\Rightarrow \displaystyle\sum _{m=1}^{6}\frac{sin\Big[\theta+\frac{m\pi}{4}-\Big(\theta+\frac{(m-1)\pi}{4}\Big)\Big]}{sin\frac{\pi}{4}\Big\\{sin\Big(\theta+\frac{(m-1)\pi}{4}sin\Big(\theta+\frac{m\pi}{4}\Big)\Big\\}}
\Rightarrow \hspace15mm \displaystyle\sum _{m=1}^{6}\frac{cot\Big(\theta+\frac{(m-1)\pi}{4}\Big)-cot\Big(\theta+\frac{m\pi}{4}\Big)}{1/\sqrt{2}}=4\sqrt{2}
\Rightarrow \hspace15mm \displaystyle\sum _{m=1}^{6} \Bigg[cot \Bigg(\theta+\frac{(m-1)\pi}{4}\Bigg)-cot\Bigg(\theta+\frac{m\pi}{4}\Bigg)\Bigg]=4
$\Rightarrow cot (\theta)-cot \Bigg(\theta+\frac{\pi}{4}\Bigg)+cot\Bigg(\theta+\frac{\pi}{4}\Bigg)-cot \Bigg(\theta+\frac{2\pi}{4}\Bigg)+...+ cot \Bigg(\theta+\frac{5\pi}{4}\Bigg)-cot \Bigg(\theta+\frac{6\pi}{4}\Bigg)=4$
\Rightarrow \hspace25mm cot \theta- cot \Bigg(\frac{3\pi}{2}+\theta\Bigg)=4
\Rightarrow \hspace35mm cot \theta +tan \theta=4
\Rightarrow \hspace25mm tan^2 \theta-4 tan \theta+1=0
\Rightarrow \hspace30mm (tan \theta-2^2)^2-3=0
\Rightarrow \hspace15mm (tan\theta-2\sqrt{3})(tan \theta-2-\sqrt{3})=0
\Rightarrow \hspace35mm tan \theta =2-\sqrt{3}
\Rightarrow \hspace35mm tan \theta =2+\sqrt{3}
\Rightarrow \hspace25mm \theta=\frac{\pi}{12}; \theta=\frac{5\pi}{12}
$\Bigg[ \because \theta \in \Bigg(0, \frac{\pi}{2}\Bigg) \Bigg]$