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

In a $\triangle ABC$,if $\angle A = 90^{\circ}$,then $\cos^{-1}\left(\frac{R}{r_2+r_3}\right)$ is equal to (in $^{\circ}$)

Given $0 \leq x \leq \frac{1}{2}$,then the value of $\tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right]$ is:

If $\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha$,where $-1 \leq x \leq 1$,$-3 \leq y \leq 3$,and $x \leq \frac{y}{3}$,then for all $x, y$,$9 x^2-6 x y \cos \alpha+y^2$ is equal to

$\cosh \left(\sinh ^{-1}(\sqrt{8})+\cosh ^{-1} 5\right)=$

If for $x \in \left(0, \frac{1}{4}\right)$,the derivative of $\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$ is $\sqrt{x} \cdot g(x)$,then $g(x)$ equals