Let $x \in (0, 1)$. The set of all $x$ such that $\sin^{-1} x > \cos^{-1} x$ is the interval

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

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 $ab < 1$.

If $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$,then $x =$

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ is equal to

If $\cot^{-1} \alpha + \cot^{-1} \beta = \cot^{-1} x$,then $x = $