The value of $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$ is

The value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$ is

If $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)$, then $\frac{d y}{d x}$ is equal to

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

The number of solutions of $\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$ where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ is equal to

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 $ab < 1$.