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)

The correct options are
C $\frac{\pi }{12}$
D $\frac{5\pi }{12}$

Given :
$\frac{1}{\sin \left(\pi /4\right)}[\frac{\sin (\theta +\pi /4-\theta )}{\sin \theta \sin (\theta +\pi /4)}+\frac{\sin (\theta +\pi /2-(\theta +\pi /4\left)\right)}{\sin (\theta +\pi /4)\cdot \sin (\theta +\pi /2)}+\dots +\frac{\sin \left(\right(\theta +3\pi /2{)}_{-}(\theta +5\pi /4))}{\sin (\theta +3\pi /2)\cdot \sin (\theta +5\pi /4)}]=4\sqrt{2}$

Using Formula for splitting numerator of first term:

$\sin (\theta +\pi /4-\theta )=\sin (\theta +\pi /4)\cos \left(\theta \right)-\cos (\theta +\pi /4)\sin \left(\theta \right)$

$\Rightarrow \sqrt{2}[\cot \theta -\cot (\theta +\pi /4)+\cot (\theta +\pi /4)-\cot (\theta +\pi /2)+\dots +\cot (\theta +5\pi /4)-\cot (\theta +3\pi /2\left)\right]=4\sqrt{2}$
$\Rightarrow \tan \theta +\cot \theta =4\Rightarrow \tan \theta =2\pm \sqrt{3}$
$\Rightarrow \theta =\frac{\pi }{12}$ or $\frac{5\pi }{12}$