If $\tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right)$,then $x =$

If $x = \sin (2 \tan^{-1} 2)$,$y = \cos (2 \tan^{-1} 3)$,and $z = \sec (2 \tan^{-1} 4)$,then:

If $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$, then $x \in$

The total number of real solutions of the equation $\theta=\tan ^{-1}(2 \tan \theta)-\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^2 \theta}\right)$ is $($Here,the inverse trigonometric functions $\sin ^{-1} x$ and $\tan ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,respectively.$)$

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = $

If $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$,where $-1 \le x \le 1$,$-2 \le y \le 2$,and $x \le \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to