For $0<\theta<\dfrac{\pi}{2}$, the solution(s) of $\displaystyle \sum_{m=1}^{6} \csc\left(\theta+\dfrac{(m-1)\pi}{4}\right), \cos\left(\theta+\dfrac{m\pi}{4}\right)=4\sqrt{2}$ is/are (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{12}$ (D) $\dfrac{5\pi}{12}$Choose the correct answer from the options given below:

Given $\displaystyle \sum_{m=1}^{6} \csc!\left(\theta+\dfrac{(m-1)\pi}{4}\right), \csc!\left(\theta+\dfrac{m\pi}{4}\right)=4\sqrt{2}$ Use the identity $\csc x \csc y=\dfrac{\cot x-\cot y}{\sin(y-x)}$ Here, $y-x=\dfrac{\pi}{4}$ and $\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$ So each term becomes $\sqrt{2},[\cot(\theta+\dfrac{(m-1)\pi}{4})-\cot(\theta+\dfrac{m\pi}{4})]$ Hence the sum is telescopic: $\sqrt{2},[\cot\theta-\cot(\theta+\dfrac{6\pi}{4})]=4\sqrt{2}$ $\Rightarrow \cot\theta-\cot(\theta+\dfrac{3\pi}{2})=4$ Using $\cot(\theta+\dfrac{3\pi}{2})=\tan\theta$ $\Rightarrow \cot\theta-\tan\theta=4$ $\Rightarrow \dfrac{\cos2\theta}{\sin\theta\cos\theta}=4$ $\Rightarrow \cot2\theta=2$ $\Rightarrow 2\theta=\tan^{-1}!\left(\dfrac{1}{2}\right)$ $\Rightarrow \theta=\dfrac{\pi}{12},\ \dfrac{5\pi}{12}$