If $x=\sin \left(2{\tan }^{-1}2\right),y=\sin \left(\frac{1}{2}{\tan }^{-1}\frac{4}{3}\right),$  then

$x=\sin \left(2{\tan }^{-1}2\right), \mathrm{Let} {\tan }^{-1}2=\theta$

$\therefore \tan \theta =2$

If  $x=\sin 2\theta =2\sin \theta \cos \theta =2\left(\frac{2}{\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right)=\frac{4}{5}$

$y=\sin \left(\frac{1}{2}{\tan }^{-1}\frac{4}{3}\right), \mathrm{Let} \varphi ={\tan }^{-1}\frac{4}{3}$

$y=\sin \frac{\varphi }{2};y>0,\tan \varphi =\frac{4}{3}$
 
$y^2={\sin }^2\frac{\varphi }{2}=\frac{1-\cos \varphi }{2}$

$y^2=\frac{1-\left({\displaystyle \frac{3}{5}}\right)}{2}=\frac{1}{5}$

$1-x=1-\frac{4}{5}=\frac{1}{5}$

$\Rightarrow y^2=1-x$