Let $\Phi: U := (1, \infty) \times (-\pi, \pi) \times (0, \infty) \to \mathbb{R}^3$ be defined as $$\Phi(r,s,t) = (rt\cos(s), rt\sin(s), \sqrt{r^2-1}).$$
I have determined that $\Phi(U) = \mathbb{R}^3\setminus N$ where $N = \{(x,y,z) \in \mathbb{R}^3\; |\; x \leq 0,\; y = 0\;, z \leq 0 \}$ and $\det \mathrm{D}\Phi$ is invertible for all $(r,s,t) \in U$. Now I'm having trouble finding the inverse of $\Phi': U \to \Phi(U)$. We have: $$rt\cos(s) = x\\ rt\sin(s) = y\\ \sqrt{r^2-1} = z$$ Solving the last equation for $r$ gives us $r = \sqrt{z^2+1}$. Plugging this into the other equations gives $t = \frac{y}{r\sin(s)}$ and $s = \arctan{\frac{y}{x}}$ (assuming $y \neq 0 \neq x$). Now if $x = 0$ we have $s = \pm\frac{\pi}{2}$ and $t = \pm\frac{y}{r}$. If $x \neq 0$ and $y = 0$, we have $s = 0$ and thus $t = \frac{x}{r}$. Now this function doesn't map $y$ to $(-\pi, \pi)$. I'm thankful for every advice!