Question

If the functions $f\left(x\right)={\sin }^{-1}\left(\frac{2x}{1+x^2}\right);g\left(x\right)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ are identical, then

Answer 1

$f\left(x\right)=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$

$\therefore f\left(x\right)=\{\begin{array}{cc}-\pi -2ta{n}^{-1}x,& x<-1\\ 2ta{n}^{-1}x,& -1\le x\le 1\\ \pi -2ta{n}^{-1}x,& x>1\end{array}$

$g\left(x\right)=co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$\therefore g\left(x\right)=\{\begin{array}{cc}-2ta{n}^{-1}x,& x<0\\ 2ta{n}^{-1}x,& x\ge 0\end{array}$

$\therefore f\left(x\right)\&g\left(x\right)$ are equal when

$x\in [0,1]$

$f\left(x\right)\&g\left(x\right)$ are equal when

Range $\in [0,\frac{\pi }{2}]$