Question

$sin\left\{2ta{n}^{-1}\sqrt{\frac{1-x}{1+x}}\right\}$

Answer 1

$y=sin\left\{2ta{n}^{-1}\sqrt{\frac{1-x}{1+x}}\right\}$
$Letx=sin\theta$
$\frac{1-x}{1+x}=\frac{1-sin\theta }{1+sin\theta }=\frac{si{n}^2\frac{\theta }{2}+co{s}^2\frac{\theta }{2}-2sin\frac{\theta }{2}.cos\frac{\theta }{2}}{si{n}^2\frac{\theta }{2}+co{s}^2\frac{\theta }{2}+2sin\frac{\theta }{2}cos\frac{\theta }{2}}$
$\frac{1-x}{1+x}=\frac{{(sin\frac{\theta }{2}-cos\frac{\theta }{2})}^2}{{(sin\frac{\theta }{2}+cos\frac{\theta }{2})}^2}$
$\therefore \sqrt{\frac{1-x}{1+x}}=\frac{sin\frac{\theta }{2}-cos\frac{\theta }{2}}{sin\frac{\theta }{2}+cos\frac{\theta }{2}}=\frac{-(1-tan\frac{\theta }{2})}{1+tan\frac{\theta }{2}}$
$\sqrt{\frac{1-x}{1+x}}=-tan(\frac{\theta }{2}-\frac{\pi }{4})=tan(\frac{\pi }{4}-\frac{\theta }{2})$
$y=sin\left[2ta{n}^{-1}\sqrt{\frac{1-x}{1+x}}\right]=sin\left[2ta{n}^{-1}\left[tan(\frac{\pi }{4}-\frac{\theta }{2})\right]\right]$
$\therefore y=sin\left[2(\frac{\pi }{4}-\frac{\theta }{2})\right]$
$\therefore y=sin(\frac{\pi }{2}-\theta )$
$\therefore y=cos\theta$
$\therefore y=cos\left(si{n}^{-1}x\right)$
$\therefore y=\sqrt{1-x^2}$