If $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$,then $x$ is equal to

Evaluate: ${\tan ^{ - 1}}\left[ {\frac{{\cos x}}{{1 + \sin x}}} \right]$

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?

$\sec ^2(\tan ^{-1} 2)+\operatorname{cosec}^2(\cot ^{-1} 3) = $ . . . . . . .

$\tan ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}\right] + \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]$ is equal to

Evaluate: $\cot ^{ - 1}\left(\frac{xy + 1}{x - y}\right) + \cot ^{ - 1}\left(\frac{yz + 1}{y - z}\right) + \cot ^{ - 1}\left(\frac{zx + 1}{z - x}\right)$