If ${\sin }^{-1}\left(\frac{2p}{1+p^2}\right)-{\cos }^{-1}\left(\frac{1-q^2}{1+q^2}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$  then prove that $x=\frac{p-q}{1+pq}$

Given 

${\sin }^{-1}\left(\frac{2p}{1+p^2}\right)-{\cos }^{-1}\left(\frac{1-q^2}{1+q^2}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

Let $p=\tan A\to A={\tan }^{-1}p$

$q=\tan B\to B={\tan }^{-1}q$

$x=\tan C\to C={\tan }^{-1}x$

Now, ${\sin }^{-1}\left(\frac{2\tan A}{1+{\tan }^{2}A}\right)-{\cos }^{-1}\left(\frac{1-{\tan }^{2}B}{1+{\tan }^{2}B}\right)$ $={\tan }^{-1}\frac{2\tan C}{1-{\tan }^{2}C}$

${\sin }^{-1}(\sin 2A)-{\cos }^{-1}(\cos 2B)={\tan }^{-1}(\tan 2C)$

$2A-2B=2C\Rightarrow A-B=C$

${\tan }^{-1}p-{\tan }^{-1}q={\tan }^{-1}x$

${\tan }^{-1}\left(\frac{p-q}{1+pq}\right)={\tan }^{-1}x$ 

$\because$$x=\frac{p-q}{1+pq}$