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.


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