$\cos \frac{{2\pi }}{{15}}\cos \frac{{4\pi }}{{15}}\cos \frac{{8\pi }}{{15}}\cos \frac{{16\pi }}{{15}} =$
$1/2$
$1/4$
$1/8$
$1/16$
Prove that $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$
If $\sin A + \cos A = \sqrt 2 ,$ then ${\cos ^2}A = $
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
$\tan 3A - \tan 2A - \tan A = $
If $\tan \,(A + B) = p,\,\,\tan \,(A - B) = q,$ then the value of $\tan \,2A$ in terms of $p$ and $q$ is