Domain of the inverse of the function given by $f(x)=\frac{1-x^2}{1+x^2}$ I want to find the domain of the inverse of this function: $$f(x):=\frac{1-x^2}{1+x^2}$$But i need a different method than considering the range of f as the domain of inverse of f. 
I simplified this to $$f(x)=-1+\frac{2}{1+x^2}$$How can i approach to find the domain of inverse of f using this?

You may write
$$
\frac{1-x^2}{1+x^2}=y
$$ giving
$$
x^2=\frac{1-y}{1+y},\qquad y\neq-1.
$$ Then you have to ensure that 
$$
\frac{1-y}{1+y}\geq0
$$ in order to obtain
$$
x=\sqrt{\frac{1-y}{1+y}}.
$$ Finally the domain you are looking for is

$$
y \in (-1,1].
$$

