Bijectivity of a trigonometric function 
Let $f : (-1,1)\to (-\pi/2,\pi/2)$ be the function defined by $f(x)= \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ the verify that $f$ is bijective

To check objectivity I assumed 2 variables $x$ and $y$ to be equal and so as to prove $f(x)=f(y)$. But I couldn't do so. I also wish to prove surjectivity.
 A: I guess that $f(x)= \arctan(\frac{2x}{1-x^2}).$ I am right ?
Now, if $x,y \in (-1,1)$ and $f(x)=f(y)$, then we have to show that $x=y.$ We get $\frac{2x}{1-x^2}=\frac{2y}{1-y^2}, $ since $\arctan$ is strictly increasing.
This gives $x-y=xy(y-x)$. Now suppose that $x \ne y.$ Then we have $xy=-1$. But this is impossible, since  $x,y \in (-1,1)$, hence $x=y.$
A: Let
$$
f(x)=\arctan\frac{2x}{1-x^2},\qquad g(x)=\frac{2x}{1-x^2}
$$
The derivative is
$$
f'(x)=\frac{1}{1+g(x)^2}g'(x)
$$
Now
$$
1+g(x)^2=1+\frac{4x^2}{(1-x^2)^2}=\frac{(1+x^2)^2}{(1-x^2)^2}
$$
and
$$
g'(x)=2\frac{1-x^2+2x^2}{(1-x^2)^2}
$$
which means that
$$
f'(x)=\frac{(1-x^2)^2}{(1+x^2)^2}\frac{2(1+x^2)}{(1-x^2)^2}=\frac{2}{1+x^2}
$$
which is positive. Since
$$
\lim_{x\to-1^+}f(x)=-\frac{\pi}{2},\qquad\lim_{x\to-1^-}f(x)=\frac{\pi}{2}
$$
you're done.
By the way, this shows that, for $x\ne\pm1$,
$$
f(x)=\begin{cases}
c_++2\arctan x & x>1 \\[4px]
c_0+2\arctan x & -1<x<1 \\[4px]
c_-+2\arctan x & x<-1
\end{cases}
$$
Then $c_0=0$, by evaluating at $0$. Since the limits of $f$ at $\pm\infty$ are $0$, we get that $c_+=-\pi$ and $c_-=\pi$.
A: $$\forall x\in(-1,1):\left(\frac x{1-x^2}\right)'=\frac{1+x^2}{(1-x^2)^2}>1$$
and
$$\forall t:(\arctan(t))'=\frac1{1+t^2}>0.$$
Both functions are monotonous and continous.
