For 0

Question

(B) Given the expression for $0 < 	heta < \frac{\pi}{2}$:$\sum_{m=1}^6 \operatorname{cosec}\left(	heta+\frac{(m-1) \pi}{4}ight) \operatorname{cose

Vedclass · Accepted Answer

$(C, D)$

Vedclass · Answer

(B) Given the expression for $0 $\sum_{m=1}^6 \operatorname{cosec}\left(	heta+\frac{(m-1) \pi}{4}ight) \operatorname{cosec}\left(	heta+\frac{m \pi}{4}ight) = 4 \sqrt{2}$Using the identity $\operatorname{cosec} A \operatorname{cosec} B = \frac{\sin(B-A)}{\sin(B-A) \sin A \sin B} = \frac{\cot A - \cot B}{\sin(B-A)}$,where $B-A = \frac{\pi}{4}$:$\sum_{m=1}^6 \frac{\cot \left(	heta+\frac{(m-1) \pi}{4}ight) - \cot \left(	heta+\frac{m \pi}{4}ight)}{\sin(\pi/4)} = 4 \sqrt{2}$Since $\sin(\pi/4) = \frac{1}{\sqrt{2}}$,we have:$\sqrt{2} \sum_{m=1}^6 \left[ \cot \left(	heta+\frac{(m-1) \pi}{4}ight) - \cot \left(	heta+\frac{m \pi}{4}ight) ight] = 4 \sqrt{2}$$\sum_{m=1}^6 \left[ \cot \left(	heta+\frac{(m-1) \pi}{4}ight) - \cot \left(	heta+\frac{m \pi}{4}ight) ight] = 4$This is a telescoping sum:$\cot 	heta - \cot \left(	heta + \frac{6\pi}{4}ight) = 4$$\cot 	heta - \cot \left(	heta + \frac{3\pi}{2}ight) = 4$$\cot 	heta + 	an 	heta = 4$$\frac{1}{	an 	heta} + 	an 	heta = 4 \Rightarrow 	an^2 	heta - 4 	an 	heta + 1 = 0$Solving for $	an 	heta$ using the quadratic formula:$	an 	heta = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$For $	an 	heta = 2 - \sqrt{3}$,$	heta = \frac{\pi}{12}$.For $	an 	heta = 2 + \sqrt{3}$,$	heta = \frac{5\pi}{12}$.Both values lie in the interval $(0, \frac{\pi}{2})$. Thus,the solutions are $\frac{\pi}{12}$ and $\frac{5\pi}{12}$.

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):

Similar Questions

If $\cos x = \frac{2 \cos y - 1}{2 - \cos y}$ where $x, y \in (0, \pi)$,then $\tan(x/2) \cot(y/2) =$

The number of $x \in [0, 2\pi]$ for which $|\sqrt{2 \sin^4 x + 18 \cos^2 x} - \sqrt{2 \cos^4 x + 18 \sin^2 x}| = 1$ is

If $0 < \theta < \frac{\pi}{2}$ and $\sin \theta \cos \theta = \frac{12}{25}$,then $\sin^4 \theta + \cos^4 \theta$ is equal to

If $\tan (\pi \cos \theta) = \cot (\pi \sin \theta)$,then $\sin \left(\frac{\pi}{4} + \theta\right) =$

If $\sin (\theta-\alpha), \sin \theta$ and $\sin (\theta+\alpha)$ are in $H.P.$,then the value of $\cos 2 \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module