If $\sin ^{-1}\left(\frac{x}{5}\right) + \csc ^{-1}\left(\frac{5}{4}\right) = \frac{\pi}{2}$,then $x = $

Let $S = \{x \in R : 0 < x < 1 \text{ and } 2 \tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\}$. If $n(S)$ denotes the number of elements in $S$,then:

Prove $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x$,where $-\frac{1}{\sqrt{2}} \leq x \leq 1$.

$\tan^{-1}(-2) - \tan^{-1}(3)$ is equal to

Considering only the principal values of the inverse trigonometric function,the value of $\tan \left(\cos ^{-1} \frac{1}{5 \sqrt{2}}-\sin ^{-1} \frac{4}{\sqrt{17}}\right)$ is

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