Prove that $f(x)=\log\sqrt{\frac{1+x}{1-x}}$ is surjective from $(-1,1)$ to $\mathbb{R}$. I have to prove that the function $\;f:(-1,1)\to \mathbb{R}\;$ defined by $f(x)=\log\sqrt{\frac{1+x}{1-x}}\;$ is bijective.
I have already proved that it is injective:
$$f(x)=f(y)$$
$$\log\sqrt{\frac{1+x}{1-x}}=\log\sqrt{\frac{1+y}{1-y}}$$
$$\log\sqrt{\frac{(1+x)(1-y)}{(1-x)(1+y)}}=0$$
$$\frac{1+x-y-xy}{1-x+y-xy}=1$$
$$x=y$$
But now, how can I prove that the function is surjective?

Let $y\in\mathbb{R}$. Then the equation
$$
y=\log\sqrt{\frac{1+x}{1-x}}
$$
becomes
$$
\frac{1+x}{1-x}=e^{2y}
$$
that solves as
$$
x=\frac{e^{2y}-1}{e^{2y}+1}
$$
This can be rewritten as $x=\tanh y$, but is not relevant. Note that
$$
-1<\frac{e^{2y}-1}{e^{2y}+1}<1
$$
if and only if
$$
-e^{2y}-1<e^{2y}-1<e^{2y}+1
$$
and both inequalities are obviously true for every $y$.
This can be simplified if you know about hyperbolic function: you need to prove that $-1<\tanh y<1$, that is, $\tanh^2y<1$ or $\sinh^2y<\cosh^2y$, which is clear because $\sinh^2y=\cosh^2y-1$.
The proof of injectivity can be shortened by noticing that both the logarithm and the square root are injective, so you just have to prove that
$$
\frac{1+x}{1-x}=\frac{1+y}{1-y}
$$
implies $x=y$. The given equality becomes
$$
1+x-y-xy=1+y-x-xy
$$
that's exactly $x=y$.
