If $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$, then $x=$

$\cos \left[ {{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{2} \right] = $

If $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$ and $\beta = \tan^{-1}\left(\frac{1}{3}\right)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\alpha - \beta$ is equal to

$\tan ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}\right] + \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]$ is equal to

Let $f(\theta) = \sin ( \tan ^{-1} ( \frac{\sin \theta}{\sqrt{\cos 2 \theta}} ) )$,where $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$,then the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is

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