Solution
= Solution
The <hyperbolic law of sines> is
$$
\frac{\sinh a}{\sin\alpha}
=\frac{\sinh b}{\sin\beta}
=\frac{\sinh c}{\sin\gamma}.
$$
Set $\gamma=\pi/2$ in the cosine laws for sides and angles and eliminate $\cos\alpha$ and $\cos\beta$. Straight algebra gives the right-triangle identities
$$
\boxed{
\tan\alpha=\frac{\sinh a}{\cosh a\sinh b},
\qquad
\tan\alpha\tan\beta\cosh c=1}.
$$