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

If $p, q, r$ are in $G.P.$ and $\tan^{-1} p, \tan^{-1} q, \tan^{-1} r$ are in $A.P.$,then $p, q, r$ satisfy the relation:

Considering the principal values of the inverse trigonometric functions,the sum of all the solutions of the equation $\cos ^{-1}(x) - 2 \sin ^{-1}(x) = \cos ^{-1}(2x)$ is equal to.

Match List $I$ with List $II$ and select the correct answer using the code given below the lists:

List $I$	List $II$
$P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value	$1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$
$Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is	$2$. $\sqrt{2}$
$R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is	$3$. $\frac{1}{2}$
$S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is	$4$. $1$

Codes: $P \quad Q \quad R \quad S$

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} = $

The sum $\sum\limits_{n = 1}^\infty {{\cot }^{ - 1}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to