Consider a regular triangle in the hyperbolic plane $\mathbb{H}^2$ (i.e. a triangle with all sides length $\alpha$ and all angles $a$).
Prove that $\cosh\alpha = \frac{\cos a}{1-\cos a}$.
I think I have to use the hyperbolic law of cosines: $$\cosh\alpha = \cosh\beta \cosh\gamma - \sinh\beta\sinh\gamma\cos a$$ simplifies to $$\cosh\alpha=\cosh^2\alpha - \sinh^2\alpha\cos a$$ in this case, but I don't see how to proceed from there.
$$\cosh\alpha = \cosh\beta \cosh\gamma - \sinh\beta\sinh\gamma\cos \alpha$$
Let
$$ \cos(\text{any of the 3 vertex angles} )=c, \cosh\alpha =ch,~\sinh\alpha =sh,~ch^2-sh^2==1 ~$$
$$ ch = ch^2- (ch^2-1) ~c \tag 1 $$
As @Blue mentions this is factorized:
$$ ch^2 (1-c) -ch + c=0 ~ \to ( ch-1)( ch-\frac{c}{1-c})=0;~ \cosh(\alpha)=(1,\frac{c}{1-c}) \tag 2 $$
Poincaré disc model.
The first solution has non-zero $(\alpha,\beta,\gamma)=\pi/3$ blunt equilateral triangle inside the boundary. Hyperbolic length of each side is finite
The second solution has hyperbolic equilateral triangle spikes touch each other $(\alpha,\beta,\gamma)= 0 $ and are perpendicular to the boundary / horizon. Hyperbolic lengths from vertex to vertex ( on the boundary ) are infinite.