Injectivity and range of $\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ To prove that $f(x)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ is injective I have tried the following:
According to the definition, for a function to be injective $f(a)=f(b) \to a=b$ for all $a,b \in D_f$. Using this I get:
$$\arctan\left(\sqrt{\frac{1+a}{1-a}}\right)=\arctan\left(\sqrt{\frac{1+b}{1-b}}\right)$$
$$\iff\left(\sqrt{\frac{1+a}{1-a}}\right)=\left(\sqrt{\frac{1+b}{1-b}}\right)$$
$$\iff\frac{1+a}{1-a}=\frac{1+b}{1-b}$$
$$\iff2a=2b$$
$$a=b$$
Hence the function is injective. (?) Now this could be total bs but I also am not able to find the proper range for this one. WolframAlpha says $0\le y<\frac{\pi }{2}$. But how does one come up with this range?
 A: The domain is $-1\leq x<1$, $\lim\limits_{x\rightarrow1^-}f(x)=\frac{\pi}{2}$, $f(-1)=0$ and $f$ is a continuous function.
Thus, the range is $\left[0,\frac{\pi}{2}\right).$
A: We need $\dfrac{1+x}{1-x}\ge0$
$\dfrac{1+x}{1-x}=0\implies x=-1 $
$\dfrac{1+x}{1-x}>0\implies-1<x<1$
WLOG $\arccos x=2y,x=\cos2y$ where $0<2y\le\pi$ using Principal values.
$$\arctan\sqrt{\dfrac{1+x}{1-x}}=\arctan\cot y=\arctan\left(\tan\left(\dfrac\pi2-y\right)\right)=\dfrac\pi2-\dfrac{\arccos x}2$$ 
Now I hope that things should be managed more easily than in the original form. 
A: The proof of injectivity is good. In order to find the range, consider the equation
$$
y=\arctan\sqrt{\frac{1+x}{1-x}}
$$
which is equivalent to $y\in(-\pi/2,\pi/2)$ and
$$
\sqrt{\frac{1+x}{1-x}}=\tan y
$$
In particular, $y\in[0,\pi/2)$, because the left-hand side is nonnegative. We can square and get
$$
\frac{1+x}{1-x}=\tan^2y
$$
that reduces to
$$
x=\frac{\tan^2y-1}{\tan^2y+1}=\sin^2y-\cos^2y=-\cos2y
$$
For every $y\in[0,\pi/2)$, we have $-1\le-\cos2y<1$, so the range is $[0,\pi/2)$.
