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?

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 $\left[0,\frac{\pi}{2}\right).$
