The expression $\frac{{{{\tan }^2}20^\circ  - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ simplifies to

$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to

$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}$ is equal to$......$.

Suppose $\theta $ and $\phi  (\ne 0)$ are such that $sec\,(\theta  + \phi ),$ $sec\,\theta $ and $sec\,(\theta  - \phi )$ are in $A.P.$ If $cos\,\theta  = k\,cos\,( \frac {\phi }{2})$ for some $k,$ then $k$ is equal to

Let $0 < x < \frac{\pi }{4}.$ Then $\sec 2x - \tan 2x = $

If $\sin A = n\sin B,$ then $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $