Proof of an identity that relates hyperbolic trigonometric function to an expression with euclidean trigonometric functions. 
Given a line $r$ and a (superior) semicircle perpendicular to $r$, and an arc $[AB]$ in the semicircle, I need to prove that
$$
\sinh(m(AB)) = \frac{\cos(\alpha)+\cos(\beta)}{\sin(\alpha)\sin(\beta)} \\
\cosh(m(AB)) = \frac{1 + \cos(\alpha)\cos(\beta)}{\sin(\alpha)\sin(\beta)}
$$
where $\alpha$ is $\angle A'OA$ and $\beta$ is $\angle B'OB$.
The argument $m(AB)$ of $sinh(m(AB))$ and $cosh(m(AB))$ above is the hyperbolic Cayley-Klein hyperbolic metric of a hyperbolic segment
$$
m(AB) = \ln \left| \frac{AA' \cdot BB'}{BA' \cdot AB'} \right|
$$
where $AB$ is the euclidean measure.
I tried writing the left side of both of them as the exponential definition. But it is so hard to manipulate the right side because always appear other segments like AO, BO and AB...
Thanks. 

Using the definition of the hyperbolic sine and hyperbolic cosine functions, we have
$$
\sinh m(AB)
 = \frac{e^{m(AB)} - e^{-m(AB)}}{2}
 = \frac{(AA'\cdot BB')^2 - (BA'\cdot AB')^2}{2(AA'\cdot BB')(BA'\cdot AB')} \\
\cosh m(AB)
 = \frac{e^{m(AB)} + e^{-m(AB)}}{2}
 = \frac{(AA'\cdot BB')^2 + (BA'\cdot AB')^2}{2(AA'\cdot BB')(BA'\cdot AB')} \\
$$
Using the law of cosines and letting $R$ be the radius of the circle, we have
$$
\begin{eqnarray}
AA'^2 &=& 2R^2(1 - \cos \alpha) \\
BB'^2 &=& 2R^2(1 - \cos \beta) \\
AB'^2 &=& 2R^2(1 + \cos \alpha) \\
BA'^2 &=& 2R^2(1 + \cos \beta) \\
\end{eqnarray}
$$
Combining these for convenience, we have
$$
\begin{eqnarray}
(AA' \cdot BB')^2 &=& 4R^4(1 - \cos \alpha - \cos \beta + \cos \alpha \cos \beta) \\
(BA' \cdot AB')^2 &=& 4R^4(1 + \cos \alpha + \cos \beta + \cos \alpha \cos \beta) \\
(AA' \cdot BB')^2 - (BA' \cdot AB')^2 &=& 8R^4(\cos \alpha + \cos \beta) \\
(AA' \cdot BB')^2 + (BA' \cdot AB')^2 &=& 8R^4(1 + \cos \alpha \cos \beta) \\
(AA' \cdot BB')(BA' \cdot AB') &=& 4R^4 \sin \alpha \sin \beta
\end{eqnarray}
$$
Plugging these back into the hyperbolic sine and hyperbolic cosine functions, we finally have
$$
\sinh m(AB) = \frac{8R^4(\cos \alpha + \cos \beta)}{2 \cdot 4R^4 \sin \alpha \sin \beta} = \frac{\cos \alpha + \cos \beta}{\sin \alpha \sin \beta} \\
\cosh m(AB) = \frac{8R^4(1 + \cos \alpha \cos \beta)}{2 \cdot 4R^4 \sin \alpha \sin \beta} = \frac{1 + \cos \alpha \cos \beta}{\sin \alpha \sin \beta} \\
$$
Which is what we wanted to prove.
