If $y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$,then $\frac{dy}{dx} = $

The number of real roots of the equation $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$ that are greater than or equal to $1$ are

$\sin ^{-1}(\sin 100) + \cos ^{-1}(\cos 100) + \tan ^{-1}(\tan 100) + \cot ^{-1}(\cot 100)$ equals to:

Identify the pair$(s)$ of functions which are identical.

If $p$ and $q$ are roots of $6x^2 + 10x + 1 = 0$,then the value of $[\tan^{-1} p + \tan^{-1} q]$ is: {where $[x]$ denotes the greatest integer less than or equal to $x$}

If $y = \operatorname{Sin}^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$ and $\frac{-3\pi}{2} < x < \frac{-\pi}{2}$,then $\frac{dy}{dx} = $