Proving a map between the triangle and the square is bijective Prove that
$$(u,v) \mapsto \left(\frac{\sin u}{\cos v}, \frac{\sin v}{\cos u}\right)=\left(x, y\right)$$
is a bijection between the interior of the triangle $T:= \{0\le u,v; u+v \le \frac {\pi}2\}$ and the square $S:=\{0\le u,v \le 1\}$.
I managed to partly show injectivity in the following way: assume
$$\begin{aligned} \frac {\sin u}{\cos v} &= \frac {\sin u'}{\cos v'} \\
 \frac {\sin v}{\cos u} &= \frac {\sin v'}{\cos u'} \end{aligned}$$
then it follows that $$\begin{aligned} \sin u\cos v' - \sin v'\cos u &= \sin u' \cos v - \sin v \cos u' \\
\sin(u-v')&=\sin(u'-v)\\
\Rightarrow u+v &=u'+v'\end{aligned}$$
Hence two points in the triangle are mapped to the same point in the square only if they lie on a straight line parallel to the hypotenuse.
I also computed the jacobian of this map and I know it is equal to $1-x^2y^2$, so that the map is smooth in the interior of the triangle. I'm sure I can use this at some point in the proof but I don't know where.
 A: A partial answer: 
To finish showing injectivity, consider a line parallel to the diagonal, parameterized as 
$$
c(t) = (t, K-t)
$$ 
for some $K$ between $0$ and $2 \pi$, and $0 \le t \le K$. 
If your alleged bijection is called $F$, let's calculate $F\circ c(t)$:
$$
F(c(t)) = \left(\frac{\sin t}{\cos (K-t)}, \frac{\sin (K-t)}{\cos t} \right)\
$$
Let's compute the derivative of the first term wrt $t$:
$$
\left(\frac{\sin t}{\cos (K-t)}\right)'
= \frac{\cos(t)\cos(K-t) - \sin(t)\sin(K-t)} {\cos^2 (K-t)}\\
= \frac{\cos(t + (K-t)} {\cos^2 (K-t)}\\
= \frac{\cos(K)} {\cos^2 (K-t)}.
$$
The numerator is a nonzero constant (for $K < \pi/2$) and the denominator is a decreasing function of $t$. Therefore the first component of $F \circ c$ is injective. 
For surjectivity, you might want to show that the boundary of the triangle maps to the boundary of the square nicely, and that the interior of the triangle maps to the interior of the square, and then note that both the triangle and square are homeomorphic to the unit disk. Then you can invoke a theorem from topology: it's impossible to map an $n$-dimensional cube onto a proper subset of itself by a continuous deformation in which the boundary remains pointwise fixed (page 1 of Alexandroff, Elementary Concepts of Topology). 
