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)

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

Let,$S=\left\{\theta \in[0,2 \pi]: 8^{2 \sin ^{2} \theta}+8^{2 \cos ^{2} \theta}=16\right\}$. Then $n ( S )+\sum_{\theta \in S}\left(\sec \left(\frac{\pi}{4}+2 \theta\right) \operatorname{cosec}\left(\frac{\pi}{4}+2 \theta\right)\right)$ is equal to.

If $\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ is the solution of $4 \cos \theta+5 \sin \theta=1$, then the value of $\tan \alpha$ is

The number of integral value $(s)$ of $'p'$ for which the equation $99\cos 2\theta  – 20\sin 2\theta  = 20p + 35$ , will have a solution is 

The expression $(1 + \tan x + {\tan ^2}x)$ $(1 – \cot x + {\cot ^2}x)$ has the positive values for $x$, given by