Proving $G(s,t)=(\cosh s \cdot \cos t, \sinh s \cdot \sin t)$ has global inverse in a certain region 
Let $G(s,t)=(\cosh s \cdot \cos t, \sinh s \cdot \sin t)$, and let $A=\{(s,t)\in\mathbb{R^2}|s>0, 0<t<2\pi\}$.
  Prove that $G_{|A}$ has global inverse and give a graphical representation of $G(A)$.

After verifying that $G_{|A}\in C^1$ and $\det(JG_{|A})\neq 0\ \ \forall (s,t)\in A$, I want to prove that $G_{|A}$ is injective. I start with $\cosh \tilde s \cdot \cos \tilde t = \cosh s \cdot \cos t$ and $\sinh \tilde s \cdot \sin \tilde t = \sinh s \cdot \sin t$, but I don't know how to prove that $\tilde s = s$ and $\tilde t = t$.
$\cosh s$ and $\sinh s$ are injective in $A$, but I don't know if this helps.
And what's the idea behind the graph I am asked to draw? Thanks in advance!
$\textbf{Attempt}$
Let $u=\cosh s \cdot \cos t,\ t=\arccos(\frac{u}{\cosh s})$
$v = \sinh s \cdot \sin t=\sinh s\cdot \sin(\arccos(\frac{u}{\cosh s}))=\sinh s\cdot\sqrt{1-(\frac{u}{\cosh s})^2}$
Given $u$, we can prove $v=g_u(s)$ is injective by verifying if its first derivative is strictily monotonic,
$g_u(s)'=\cosh s\cdot \sqrt{1-(\frac{u}{\cosh s})^2}+\sinh s\cdot\frac{1}{2\sqrt{1-(\frac{u}{\cosh s})^2}}\cdot2\frac{u}{\cosh s}\cdot\frac{u}{\cosh^2s}\cdot \sinh s=$
$=\cosh s\cdot \sqrt{1-(\frac{u}{\cosh s})^2}+\frac{u^2\sinh^2 s}{\cosh^3\sqrt{1-(\frac{u}{\cosh s})^2}}>0$, because $\cosh s>1,\ \sinh s>0\ \forall s>0,$
and $1-(\frac{u}{\cosh s})^2=1-\frac{\cosh^2s\cdot\cos^2 t}{\cosh^2s}=1-\cos^2t=\sin^2t\in[0,1]\ \forall t\in(0,2\pi)$
But $\sin t=0$ if $t=\pi$, so the proof might fail here.
Anyway, we can repeat the same prodecure to prove that $u$ is injective. From $v$, we have that: $t=\arcsin(\frac{v}{\sinh s})\to u=\cosh s\cdot \cos(\arcsin(\frac{v}{\sinh s}))=\cosh s\cdot\sqrt{1-(\frac{v}{\sinh s})^2}$
Given $v$, we prove $u=h_v(s)$ is injective:
$h_v(s)'= \sinh s\cdot\sqrt{1-(\frac{v}{\sinh s})^2}+\cosh s\cdot\frac{1}{2\sqrt{1-(\frac{v}{\sinh s})^2}}\cdot2\frac{v}{\sinh s}\cdot\frac{v}{\sinh^2s}\cdot \cosh s=$
$=\sinh s\cdot \sqrt{1-(\frac{v}{\sinh s})^2}+\frac{v^2\cosh^2 s}{\sinh^3\sqrt{1-(\frac{v}{\sinh s})^2}}>0$, because $\cosh s>1,\ \sinh s>0\ \forall s>0,$
and $1-(\frac{v}{\sinh s})^2=1-\frac{\sinh^2s\cdot\sin^2 t}{\sinh^2s}=1-\sin^2t=\cos^2t\in[0,1]\ \forall t\in(0,2\pi)$
And we have a similar problem: $\cos t=0$ if $t=\pi/2$.
What's wrong in my attempt? Am I making wrong assumptions?
 A: As for the graphic representation note that for $u=\cosh s\cdot\cos t$, $v=\sinh s\cdot\sin t$: 
$$ \frac{u^2}{\cosh^2 s}+\frac{v^2}{\sinh^2 s} = 1$$
$$ \frac{u^2}{\cos^2 t}-\frac{v^2}{\sin^2 t} = 1$$
Which means that the lines of constant $s$ are ellipses, and the lines of constant $t$ are hyperbolae. You can also check that they will always intersect at right angles. I believe that for the graphical representation of $G$ you're supposed to sketch the image of a grid under the action of $G$.
A: Hint Under the usual identification $\Bbb R^2 \leftrightarrow \Bbb C$, $(u, v) \leftrightarrow u + iv$, the transformation $G$ is
$$G(s + it) = \cosh s \cos t + i \sinh s \sin t ,$$
but writing $z := s + it$ this is precisely
$$G(z) = \cosh z .$$

 We need to show that $G\vert_A$ admits an inverse---which would necessarily be a branch of the inverse hyperbolic cosine---where $A$ is the open half-strip $\{\operatorname{Re} z > 0, 0 < \operatorname{Im} z < 2 \pi\}$. But by definition $G\vert_A^{-1}$ is exactly the principal branch of the inverse hyperbolic cosine, namely $$\operatorname{arcosh} w = \log(w + \sqrt{w - 1} \sqrt{w + 1}) .$$ Here, the branches of $\log$ and $\sqrt{\cdot}$ are the principal ones.

