Question

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

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

Answer 1

Answer: (C), (D)

For $0<\theta <\frac{\pi }{2}$
$\sum _{m=1}^{6}cosec(\theta +\frac{(m-1)}{4})cosec(\theta +fracm\pi 4)=4\sqrt{2}$
$\Rightarrow \sum _{m=1}^6\frac{1}{sin(\theta +\frac{(m-1)\pi }{4})sin(\theta +\frac{m\pi }{4})}=4\sqrt{2}$
$\Rightarrow \sum _{m=1}^6\frac{sin[\theta +\frac{m\pi }{4}-(\theta +\frac{(m-1)\pi }{4}\left)\right]}{sin\frac{\pi }{4}\left\{sin\right(\theta +\frac{(m-1)\pi }{4}sin(\theta +\frac{m\pi }{4})\}}$
$\Rightarrow \sum _{m=1}^6\frac{cot(\theta +\frac{(m-1)\pi }{4})-cot(\theta +\frac{m\pi }{4})}{1/\sqrt{2}}=4\sqrt{2}$
$\Rightarrow \sum _{m=1}^6\left[cot\right(\theta +\frac{(m-1)\pi }{4})-cot(\theta +\frac{m\pi }{4}\left)\right]=4$
$\Rightarrow cot(\theta )-cot(\theta +\frac{\pi }{4})+cot(\theta +\frac{\pi }{4})-cot(\theta +\frac{2\pi }{4})+...+cot(\theta +\frac{5\pi }{4})-cot(\theta +\frac{6\pi }{4})=4$
$\Rightarrow cot\theta -cot(\frac{3\pi }{2}+\theta )=4$
$\Rightarrow cot\theta +tan\theta =4$
$\Rightarrow ta{n}^2\theta -4tan\theta +1=0$
$\Rightarrow (tan\theta -2^2{)}^2-3=0$
$\Rightarrow (tan\theta -2\sqrt{3})(tan\theta -2-\sqrt{3})=0$
$\Rightarrow tan\theta =2-\sqrt{3}$
$\Rightarrow tan\theta =2+\sqrt{3}$
$\Rightarrow \theta =\frac{\pi }{12};\theta =\frac{5\pi }{12}$
$[\because \theta \in (0,\frac{\pi }{2}\left)\right]$