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

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 $\theta $ and $\phi $ are acute satisfying $\sin \theta = \frac{1}{2},$ $\cos \phi = \frac{1}{3},$ then $\theta + \phi \in $

If $\sin \theta = \sqrt 3 \cos \theta , - \pi < \theta < 0$, then $\theta = $

Let $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}$ and $\beta=\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right)$, then $\frac{1}{6}(\beta-14)^2$ is equal to

Let $S=\{x \in R: \cos (x)+\cos (\sqrt{2} x)<2\}$, then