The number of solutions of the equation $32^{\tan ^{2} x}+32^{\sec ^{2} x}=81,0 \leq x \leq \frac{\pi}{4}$ is :

If $\alpha ,\beta ,\gamma $ be the angles made by a line with $x, y$ and $z$ axes respectively so that $2\left( {\frac{{{{\tan }^2}\,\alpha }}{{1 + {{\tan }^2}\,\alpha }} + \frac{{{{\tan }^2}\,\beta }}{{1 + {{\tan }^2}\,\beta }} + \frac{{{{\tan }^2}\,\gamma }}{{1 + {{\tan }^2}\,\gamma }}} \right) = 3\,{\sec ^2}\,\frac{\theta }{2},$ then $\theta =$

The number of real solutions $x$ of the equation $\cos ^2(x \sin (2 x))+\frac{1}{1+x^2}=\cos ^2 x+\sec ^2 x$ is

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 $\cos 2\theta = (\sqrt 2 + 1)\,\,\left( {\cos \theta - \frac{1}{{\sqrt 2 }}} \right)$, then the value of $\theta $ is

Number of solutions of equation $secx = 1 + cosx + cos^2x + ........ \infty$ in $x \in [-50 \pi, 50 \pi]$ is -