Bijection from $\{ (x,y) : 0 \leq x ,y \leq 1\}$ to $\left\lbrace (u,v) : u,v \geq 0, u+v \leq \frac{\pi}{2} \right\rbrace$ Suppose the set $M :=\{ (x,y) : 0 \leq x ,y \leq 1\}$. Now we define
$$ u := \arccos \sqrt{\frac{1-x^2}{1-x^2y^2}} \ \ \ \ \ \text{and} \ \ \ \ \  v:= \arccos  \sqrt{\frac{1-y^2}{1-x^2 y^2}}.$$
How can I show, that this new coordinatens $(u,v)$ map each point  $(x,y) \in M$ bijective to a point in the new set
$$ \widetilde{M} := \left\lbrace (u,v) : u,v \geq 0, u+v \leq \frac{\pi}{2} \right\rbrace?$$
I found out that we have $x = \frac{\sin u}{\cos v}$ and $y = \frac{\sin v }{\cos u}$ but I do not understand how to get the condition $u+v \leq \frac{\pi}{2} $.
 A: One can show that
$$
  u := \arccos \sqrt{\frac{1-x^2}{1-x^2y^2}} \quad \text{and} \quad
  v:= \arccos \sqrt{\frac{1-y^2}{1-x^2 y^2}} \quad \tag 1
$$
is a bijective mapping of the  open rectangle
$M :=\{ (x,y) : 0 < x ,y < 1\}$ onto the open triangle
$\widetilde{M} := \left\lbrace (u,v) : u,v > 0, u+v < \frac{\pi}{2} \right\rbrace$.
The inverse mapping is
$$
 x := \frac{\sin u}{\cos v} \quad \text{and} \quad y := \frac{\sin v }{\cos u}  \tag 2
$$
as shown below.
Since
$$
   (u(1, y), v(1, y)) = (\frac \pi 2, 0) \\
   (u(x, 1), v(x, 1)) = (0, \frac \pi 2) \\
$$
the mapping cannot be extended one-to-one to the boundary of the square $M$.
Now let $(x, y) \in M$ and define $(u,v)$ by $(1)$. It is clear that
both $u$ and $v$ are in the range $(0, \frac \pi 2)$. To show that
$u + v < \frac \pi 2$, we compute
$$
  \cos (u + v) = \cos u \cos v - \sin u \sin v = 
   \cos u \cos v - \sqrt{1 - \cos^2 u}  \sqrt{1 - \cos^2 v} \\
 = \sqrt{\frac{1-x^2}{1-x^2y^2}} \sqrt{\frac{1-y^2}{1-x^2 y^2}}
 - \sqrt{\frac{x^2(1-y^2)}{1-x^2y^2}} \sqrt{\frac{y^2(1-x^2)}{1-x^2 y^2}} \\
= (1 - xy) \sqrt{\frac{1-x^2}{1-x^2y^2}} \sqrt{\frac{1-y^2}{1-x^2 y^2}}
> 0 \quad .
$$
It follows that $u + v < \frac \pi 2$.
So up to now we know that $(1)$ is a mapping from the square $M$
into the triangle $\widetilde{M}$. To show that the mapping is in fact
bijective, we verify that $(2)$ is the inverse mapping from
$\widetilde{M}$ to $M$.
Let $(u, v) \in \widetilde{M}$ and define $(x, y)$ by $(2)$.
It is clear that both $x$ and $y$ are positive. From
$u + v < \frac \pi 2$ it follows that
$$
  \sin u < \sin(\frac \pi 2 - v) = \cos v \quad \Longrightarrow \quad x < 1 \, ,\\
  \sin v < \sin(\frac \pi 2 - u) = \cos u \quad \Longrightarrow \quad y < 1 \, .$$
and therefore $(x, y) \in M$.
To verify that $(x, y)$ is mapped back exactly to the given $(u, v)$
is a straightforward calculation and I'll leave that final part to you :)
A: From your equations:
$$\begin{gathered}
  u(x,y) = {\cos ^{ - 1}}(\sqrt {\frac{{1 - {x^2}}}{{1 - {x^2}{y^2}}}} ) \hfill \\
  v(x,y) = \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}{y^2}}}}  \hfill \\ 
\end{gathered} $$
I got:
$$\begin{gathered}
  co{s^2}(u) = 1 - {v^2} \cdot {x^2} \hfill \\
  {v^2} \cdot {x^2} = si{n^2}(u) \hfill \\
  {x^2} = \frac{{si{n^2}(u)}}{{{v^2}}} \hfill \\ 
\end{gathered} $$
and
$$\begin{gathered}
  {v^2} = \frac{{1 - {y^2}}}{{1 - \frac{{si{n^2}(u)}}{{{v^2}}}{y^2}}} \hfill \\
  {v^2} - si{n^2}(u){y^2} = 1 - {y^2} \hfill \\
  {y^2}co{s^2}(u) = 1 - {v^2} \hfill \\
  {y^2} = \frac{{1 - {v^2}}}{{co{s^2}(u)}} \hfill \\ 
\end{gathered}$$
This gives me
$$\begin{gathered}
  x(u,v) = \frac{{sin(u)}}{v} \hfill \\
  y(u,v) = \frac{{\sqrt {1 - {v^2}} }}{{cos(u)}} \hfill \\ 
\end{gathered} $$
for $u$ and $v$ in an allowed range. The condition for "allowed" is $$u + v \leqslant \frac{\pi }{2}$$
otherwise you can't do root operation.
