Question

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

Answer 1

Let ${\tan }^{-1}2=\theta$
$\Rightarrow \tan \theta =2$
$\sin 2\theta =2.\frac{2}{\sqrt{5}}.\frac{1}{\sqrt{5}}=\frac{4}{5}=x$
Substitute ${\tan }^{-1}\frac{4}{3}=\Phi$
$\tan \Phi =\frac{4}{3}$
$y=s\in \frac{\Phi }{2}\Rightarrow y^2={\sin }^2\frac{\Phi }{2}=\frac{1-\cos \Phi }{2}$
$=\frac{1-\frac{3}{5}}{2}=\frac{1}{5}$
$\Rightarrow x+y^2=\frac{4}{5}+\frac{1}{5}=1$
$\therefore x+y^2=1$