$\sin (2{\sin ^{ - 1}}0.8) = $

$\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(-\frac{1}{2}\right)}=$

Let $a, b$ and $c$ be the lengths of the sides of a triangle with its opposite angles $A, B$ and $C$ respectively. If $a=3, b=4$ and $A=\sin^{-1}\left(\frac{3}{4}\right)$,then the angle $B$ is (in $^{\circ}$)

$\tan^{-1} [2 \cos (2 \sin^{-1} \frac{1}{2})] = \dots \dots \dots$

If $\cos^{-1} x = y$, then $\dots \dots \dots$

If $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)$,then the value of $x$ is