Question

$y=co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right);0<x<1$
$x=tan\theta$
$y=co{s}^{-1}\left(cos2\theta \right)=2\theta$
$y=2ta{n}^{-1}x$

Answer 1

$y=co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
Let $x=tan\theta$
$1-x^2=1-ta{n}^2\theta$
$1+x^2=1+ta{n}^2\theta =se{c}^2\theta$
$\frac{1-x^2}{1+x^2}=\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=\frac{co{s}^2\theta -si{n}^2\theta }{se{c}^2\theta .co{s}^2\theta }$
$\therefore \frac{1-x^2}{1+x^2}=cos2\theta$
$y=co{s}^{-1}\left(cos2\theta \right)=2\theta$
But $x=tan\theta \Rightarrow =\theta =ta{n}^{-1}x$
$\therefore y=2ta{n}^{-1}x$
$\frac{dy}{dx}=2.\frac{d}{dx}\left(ta{n}^{-1}x\right)=\frac{2}{1+x^2}[\because \frac{d}{dx}(ta{n}^{-1}x)=\frac{1}{1+x^2}]$