In the figure, $\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(B E)=4(A B)$. If the area of $\triangle CAB$ is $2 \sqrt{3}-3$ unit $^2$, when $\frac{\theta_2}{\theta_1}$ is the largest, then the perimeter (in unit) of $\triangle CED$ is equal to $...........$.

If $2\tan A = 3\tan B,$ then $\frac{{\sin 2B}}{{5 - \cos 2B}}$ is equal to

The expression $[1 - sin (3\pi - \alpha ) + cos (3\pi + \alpha )]$ $\left[ {1\,\, - \,\,\sin \,\left( {\frac{{3\,\pi }}{2}\,\, - \,\,\alpha } \right)\,\, + \,\,\cos \,\left( {\frac{{5\,\pi }}{2}\,\, - \,\,\alpha } \right)} \right]$ when simplified reduces to :

The value of $\frac{1}{4} \,\,tan \frac{\pi}{8} +\frac{1}{8} \,\,tan \frac{\pi}{16}+\frac{1}{16} \,\,tan \frac{\pi}{32}+.\,.\,.\,\infty  $ terms is equal to-

If ${\rm{cosec}}\theta = \frac{{p + q}}{{p - q}},$ then $\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = $

If $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,where $0 \le \alpha ,\beta \le \frac{\pi }{4}$ . Then $\tan 2\alpha =$