$\tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) = $

Consider the following statements:
Assertion $(A)$: When $x, y, z$ are positive numbers, then $\operatorname{Tan}^{-1}\left(\sqrt{\frac{x(x+y+z)}{y z}}\right)+\operatorname{Tan}^{-1}\left(\sqrt{\frac{y(x+y+z)}{x z}}\right)+\operatorname{Tan}^{-1}\left(\sqrt{\frac{z(x+y+z)}{x y}}\right) = \pi$
Reason $(R)$: $\operatorname{Tan}^{-1} a + \operatorname{Tan}^{-1} b = \operatorname{Tan}^{-1}\left(\frac{a+b}{1-ab}\right)$ if $a > 0$ and $b > 0$ and $ab < 1$.

Find the value of $\cot \left(\tan ^{-1} a+\cot ^{-1} a\right)$.

If ${\sin ^{ - 1}}x + {\cot ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{2},$ then $x$ is

$2{\sin ^{ - 1}}\frac{3}{5} + {\cos ^{ - 1}}\frac{{24}}{{25}} = $

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