For 0

Question

For $0 < 	heta < \frac{\pi}{2}$

$\sum_{m=1}^6 \operatorname{cosec}\left(	heta+\frac{(m-1) \pi}{4}ight) \operatorname{cosec}\left(	heta+\

Vedclass · Accepted Answer

$(C,D)$

Vedclass · Answer

For $0 < 	heta < \frac{\pi}{2}$

$\sum_{m=1}^6 \operatorname{cosec}\left(	heta+\frac{(m-1) \pi}{4}ight) \operatorname{cosec}\left(	heta+\frac{m \pi}{4}ight)=4 \sqrt{2}$

$\Rightarrow \sum_{m=1}^6 \frac{1}{\sin \left(	heta+\frac{(m-1) \pi}{4}ight) \sin \left(	heta+\frac{m \pi}{4}ight)}=4 \sqrt{2}$

$\Rightarrow \sum_{m=1}^6 \frac{\sin \left[	heta+\frac{m \pi}{4}-\left(	heta+\frac{(m-1) \pi}{4}ight)ight]}{\sin \frac{\pi}{4}\left\{\sin \left(	heta+\frac{(m-1) \pi}{4}ight) \sin \left(	heta+\frac{m \pi}{4}ight)ight\}}=4 \sqrt{2}$

$\Rightarrow \sum_{m=1}^6 \frac{\cot \left(	heta+\frac{(m-1) \pi}{4}ight)-\cot \left(	heta+\frac{m \pi}{4}ight)}{1 / \sqrt{2}}=4 \sqrt{2}$

$\Rightarrow \sum_{m=1}^6\left[\cot \left(	heta+\frac{(m-1) \pi}{4}ight)-\cot \left(	heta+\frac{m \pi}{4}ight)ight]=4$

$\Rightarrow \cot (	heta)-\cot \left(	heta+\frac{\pi}{4}ight)+\cot \left(	heta+\frac{\pi}{4}ight)-\cot \left(	heta+\frac{2 \pi}{4}ight)+$

$\cdots+\cot \left(	heta+\frac{5 \pi}{4}ight)-\cot \left(	heta+\frac{6 \pi}{4}ight)=4$

$\Rightarrow $$ \cot 	heta-\cot \left(\frac{3 \pi}{2}+	hetaight)=4$

$\Rightarrow $$ \cot 	heta+	an 	heta=4 \quad \Rightarrow \quad 	an ^2 	heta-4 	an 	heta+1=0$

$\Rightarrow $$ (	an 	heta-2)^2-3=0$

$\Rightarrow $$ (	an 	heta-2+\sqrt{3})(	an 	heta-2-\sqrt{3})=0$

$\Rightarrow $$ 	an 	heta=2-\sqrt{3} 	ext { or } 	an 	heta=2+\sqrt{3}$

$\Rightarrow $$ 	heta=\frac{\pi}{12} ; 	heta=\frac{5 \pi}{12}$

$\because $$ 	heta \in\left(0, \frac{\pi}{2}ight) .$

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)

$(A)$ $\frac{\pi}{4}$ $(B)$ $\frac{\pi}{6}$ $(C)$ $\frac{\pi}{12}$ $(D)$ $\frac{5 \pi}{12}$

Similar Questions

Number of solution$(s)$ of the equation $ln(1 + sin^2x) = 1 -ln(5 + x^2)$ is -

The number of solutions of the equation $\cos \left(x+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x, x \in[-3 \pi$ $3 \pi]$ is

If $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x)2\cos x = 0$ then

If $\sin \theta + \cos \theta = \sqrt 2 \cos \alpha $, then the general value of $\theta $ is

The number of solutions to $\sin \left(\pi \sin ^2 \theta\right)+\sin \left(\pi \cos ^2 \theta\right)=2 \cos \left(\frac{\pi}{2} \cos \theta\right)$ satisfying $0 \leq \theta \leq 2 \pi$ is

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) $(A)$ $\frac{\pi}{4}$ $(B)$ $\frac{\pi}{6}$ $(C)$ $\frac{\pi}{12}$ $(D)$ $\frac{5 \pi}{12}$