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(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

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$(C, D)$

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(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):

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Assertion $(A)$: If $\sqrt{4 \sin^4 \theta + \sin^2 2\theta} + 4 \cos^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = 2$,then $\theta$ lies in the $3^{\text{rd}}$ quadrant or $4^{\text{th}}$ quadrant.
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