$\operatorname{Sin}^{-1}(-\cos 2) + \operatorname{Cos}^{-1}(\sin 3) + \operatorname{Tan}^{-1}(\cot 5) = $

Show that $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} \frac{3}{4}$

$\frac{d}{dx}[\tan^{-1}(\cot x) + \cot^{-1}(\tan x)] = $

Consider the following statements.
$I$. $\sin ^{-1}(y^2-4y+6)+\cos ^{-1}(y^2-4y+6) = \frac{\pi}{2}, \forall y \in R$
$II$. $\sec ^{-1}(y^2-4y+6)+\operatorname{cosec}^{-1}(y^2-4y+6) = \frac{\pi}{2}, \forall y \in R$
Which of the above statement$(s)$ is/are true?

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{2} = \alpha$,where $-1 \leq x \leq 1, -2 \leq y \leq 2, x \leq \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to

If $3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}$,then $x$ =