$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $

$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $

If $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ then $x = $

$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ is equal to

Let $\frac{\pi}{2} < x < \pi$ be such that $\cot x=\frac{-5}{\sqrt{11}}$. Then $\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$ is equal to

If $\sin \theta+\cos \theta=\frac{1}{2}$, then $16(\sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta))$ is equal to: