Proving a formula for the area of a hyperbolic right triangle If K is the area of a hyperbolic right triangle ABC in which the right angle is at C, prove that
$$ \sin K=\frac{ \sinh a  \sinh b}{1+\cosh a\cosh b}$$
My attempt at the solution: I basically need help getting to the expression of Area K. I know from there to take the sine of K obviously and apply some identities...but I need some guidance as to how to get to that point. 
So things I know: $ \gamma= \frac \pi 2 $
$\cosh c = \cosh a \cosh b$
$ Area = \pi - (\alpha + \beta + \gamma) $ then... $\sin K = \sin [\pi -(\alpha + \beta + \frac \pi 2)] = \sin\alpha\sin\beta-\cos\alpha\cos\beta  $
is this the right direction to be heading into?

There are many ways, so why should it not be good? 
$$\begin{align}
\sin K &= \sin \left(\pi -(\alpha + \beta + \frac{ \pi}{ 2} )\right) \\[4pt] 
&= \sin\left(\frac 12 \pi -(\alpha + \beta )\right) \\[4pt]
&= \cos(\alpha + \beta ) \\[4pt]
&= \cos\alpha\cos\beta-\sin\alpha\sin\beta \tag{1}
\end{align}$$
Then, by the trigonometry of hyperbolic right triangles, 
$$\sin\alpha = \frac{ \sinh a }{ \sinh c } \quad\sin\beta = \frac{ \sinh b }{ \sinh c} \quad \cos\alpha = \frac{ \tanh b }{ \tanh c } \quad \cos\beta = \frac{ \tanh a }{ \tanh c} \tag{2}$$
you get
$$\frac{ \tanh a \tanh b }{ \tanh^2 c}-\frac{ \sinh a \sinh b}{ \sinh^2 c } \tag{3}$$
and then
$$\frac{ \tanh a \tanh b \sinh^2 c-\sinh a \sinh b \tanh^2 c}{\sinh^2 c \tanh^2 c} \tag{4}$$
and so on.
