Let $\theta, \phi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1, \tan (2 \pi-\theta)>0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. Then $\phi$ cannot satisfy

$(A)$ $0 < \phi<\frac{\pi}{2}$ $(B)$ $\frac{\pi}{2} < \phi<\frac{4 \pi}{3}$

$(C)$ $\frac{4 \pi}{3} < \phi<\frac{3 \pi}{2}$ $(D)$ $\frac{3 \pi}{2} < \phi < 2 \pi$

If $S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| {\begin{array}{*{20}{c}}
0&{\cos {\mkern 1mu} x}&{ – \sin {\mkern 1mu} x}\\
{\sin {\mkern 1mu} x}&0&{\cos {\mkern 1mu} x}\\
{\cos {\mkern 1mu} x}&{\sin {\mkern 1mu} x}&0
\end{array}} \right| = 0} \right\},$ then $\sum\limits_{x \in S} {\tan \left( {\frac{\pi }{3} + x} \right)} $ is equal to

If $e ^{\left(\cos ^{2} x+\cos ^{4} x+\cos ^{6} x+\ldots \ldots \infty\right) \log _{e} 2}$ satisfies the equation $t ^{2}-9 t +8=0,$ then the value of $\frac{2 \sin x}{\sin x+\sqrt{3} \cos x}\left(0 < x < \frac{\pi}{2}\right)$ is

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

If $A, B, C, D$ are the angles of a cyclic quadrilateral taken in order, then
$cos(180^o + A) + cos(180^o -B) + cos(180^o -C) -sin(90^o -D)=$