Area of a right angled hyperbolic triangle as function of side lengths I was puzzeling with  Area of hyperbolic triangle definition and could not figure it out, but then i thought there should be a (maybe solvable)  simpler problem so here it is:
suppose:


*

*an hyperbolic plane with a Gaussian curvature of -1 

*on this surface there is a triangle $\triangle ABC$

*$\angle C$ is a right angle 


Then given the lengths of sides a and b what is the area?
Off course we could do:
the area $ = \frac{\pi}{2} - \angle A -\angle B$ or
$ \frac{\pi}{2} - arctan (\frac{\tanh(a)}{\sinh(b)} )- arctan (\frac{\tanh(b)}{\sinh(a)}) $  
(free after http://en.wikipedia.org/wiki/Hyperbolic_triangle )
But is there not a  nicer formula that does not contain any or only one trigonometric function?
(I guess the hyperbolic functions are needed) 
UPDATE 07/10/2015
following https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Arctangent_addition_formula
$$ \arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) $$
and 
$$ \frac{\pi}{2} -arctan (\frac{u}{v}) = arctan (\frac{v}{u})$$
(complement tangent = cotangent)
I get to
$$ Area = arctan(\frac{ \sinh(a)\sinh(b) - \tanh(a)tanh(b)}{\sinh(a)\tanh(a)+ \sinh(b)\tanh(b)} )$$
Can this be simplified any further? 
(Do i overlook the obvious again , or even did i make a mistake on the way somewhere?) 
 A: Let's consider an arbitrary hyperbolic triangle with angles $\alpha$, $\beta$, $\gamma$ opposite respective sides $a$, $b$, $c$. The area ($T$) is given by
$$T = \pi - \alpha - \beta - \gamma \tag{0}$$
so that, writing $X_2$ for $X/2$,
$$\begin{align}
\cos T_2 &= \cos\left(\frac{\pi}{2}-\alpha_2-\beta_2-\gamma_2\right) = \sin\left(\alpha_2+\beta_2+\gamma_2\right) \tag{1a}\\
&= \sin\alpha_2 \cos\beta_2\cos\gamma_2+\cos\alpha_2\sin\beta_2\cos\gamma_2 \tag{1b} \\[4pt]
&+\cos\alpha_2\cos\beta_2\sin\gamma_2-\sin\alpha_2\sin\beta_2\sin\gamma_2
\end{align}$$
The hyperbolic Law of Cosines tells us, for instance,
$$\cos\gamma = \frac{\cosh a \cosh b - \cosh c}{\sinh a \sinh b} \tag{2}$$
so that
$$\begin{align}
\cos^2\gamma_2 &= \frac{1+\cos\gamma}{2} = \frac{\cosh(a+b)-\cosh c}{2\sinh a\sinh b} = \frac{\sinh s \sinh s_c}{\sinh a\sinh b} \tag{3a} \\[4pt]
\sin^2\gamma_2 &= \frac{1-\cos\gamma}{2} = \frac{\cosh c-\cosh(a-b)}{2\sinh a\sinh b} = \frac{\sinh s_a \sinh s_b}{\sinh a\sinh b} \tag{3b}
\end{align}$$
Likewise,
$$\cos^2\alpha_2 = \frac{\sinh s \sinh s_a}{\sinh b\sinh c} \qquad 
\sin^2\alpha_2 = \frac{\sinh s_b \sinh s_c}{\sinh b\sinh c} \tag{4}$$ 
$$\cos^2\beta_2 = \frac{\sinh s \sinh s_b}{\sinh c\sinh a} \qquad \sin^2\beta_2 = \frac{\sinh s_c \sinh s_a}{\sinh c\sinh a} \tag{5}$$ 
From these, we have
$$\begin{align}
\cos T_2 
&= \frac{s_0 s_b s_c+s_0 s_c s_a+s_0 s_a s_b - s_a s_b s_c}{\sinh a\sinh b\sinh c} \tag{6a}\\[4pt]
&= \frac{2\sinh a_2 \sinh b_2\sinh c_2\;(1+\cosh a + \cosh b + \cosh c)}{8\;\sinh a_2 \cosh a_2\;\sinh b_2 \cosh b_2\;\sinh c_2 \cosh c_2}
\end{align} \tag{6b}$$
where the numerator of $(6b)$ is a bit of a slog. (See Addendum below.) Ultimately, we conclude

$$\begin{align}
\cos T_2 &= \frac{1+\cosh a + \cosh b + \cosh c}{4\;\cosh a_2\cosh b_2\cosh c_2} \tag{7a}\\[4pt]
\sin T_2 &= \frac{\sqrt{1-\cosh^2 a-\cosh^2 b - \cosh^2 c+ 2 \cosh a\cosh b\cosh c}}{4\cosh a_2 \cosh b_2 \cosh c_2} \tag{7b}
\end{align}$$


In the particular case of a right triangle $\gamma = \pi/2$, we have $\cosh c = \cosh a \cosh b$, and the above formulas reduce thusly: 
$$\begin{align}
\cos T_2 &= \frac{(1+\cosh a)(1 + \cosh b)}{4\;\cosh a_2\cosh b_2\cosh c_2} = \frac{4\cosh^2 a_2 \cosh^2 b_2}{4\cosh a_2 \cosh b_2 \cosh c_2} = \frac{\cosh a_2 \cosh b_2}{\cosh c_2} \tag{8a}\\[4pt]
\sin T_2 &= \frac{\sqrt{(\cosh^2 a-1)(\cosh^2 b-1)}}{4\cosh a_2 \cosh b_2 \cosh c_2} = \frac{\sinh a\sinh b}{4\cosh a_2 \cosh b_2 \cosh c_2} = \frac{\sinh a_2 \sinh b_2}{\cosh c_2} \tag{8b}
\end{align}$$
so that

$$\tan T_2 = \tanh a_2 \tanh b_2 \tag{9}$$


Addendum. Here's a walkthrough of deriving the numerator of $(5)$. I did this mostly on-the-fly; the reader is encouraged to seek a much quicker path. (Expanding everything in terms of exponentials is tedious, but effective. Of course, a computer algebra system can verify the relation in an instant.)
$$\begin{align}
&\phantom{+\;} \sinh s \sinh s_b \sinh s_c + \sinh s \sinh s_c \sinh s_a \tag{Aa}\\
&+ \sinh s \sinh s_a \sinh s_b - \sinh s_a \sinh s_b \sinh s_c \\[6pt]
= &\phantom{+\;} \sinh s \sinh s_c \left( \sinh s_b + \sinh s_a \right) \tag{Ab}\\
&+ \sinh s_a \sinh s_b \left( \sinh s - \sinh s_c \right) \\[6pt]
= &\phantom{+\;\;} \frac12 \left(\cosh(s\;+s_c)-\cosh(s\;-s_c)\right)\cdot 2 \sinh\frac{s_a+s_b}{2}\;\cosh\frac{s_a-s_b}{2} \tag{Ac} \\[4pt]
&+ \frac12 \left(\cosh(s_a+s_b)-\cosh(s_a-s_b)\right) \cdot 2\cosh\frac{s\;+s_c}{2}\;\sinh\frac{s\;-s_c}{2} \\[6pt]
= &\phantom{+\;} \left(\cosh(a+b)-\cosh c\right)\cdot \sinh c_2 \cosh(a_2-b_2) \tag{Ad} \\[4pt]
&+ \left(\cosh c-\cosh(a-b)\right) \cdot \cosh(a_2+b_2)\;\sinh c_2 \\[6pt]
= &\phantom{+\;}\sinh c_2 \left(\;
\begin{array}{c} (\cosh a\cosh b - \cosh c)\;( \cosh(a_2-b_2) - \cosh(a_2+b_2)) \tag{Ae} \\[2pt]
+ \sinh a \sinh b\;( \cosh( a_2 - b_2 ) + \cosh(a_2 + b_2 ) ) 
\end{array}
\;\right) \\[4pt]
= &\phantom{+\;}\sinh c_2 \left(\;
\begin{array}{c} 
-2 \sinh a_2 \sinh b_2 (\cosh a\cosh b - \cosh c) \\[2pt]
+ 2 \sinh a \sinh b \cosh a_2 \cosh b_2
\end{array}
\;\right) \tag{Af} \\[4pt]
= &\phantom{+\;}2 \sinh c_2 \left(\;
\begin{array}{c} 
\sinh a_2 \sinh b_2 (\cosh c - \cosh a\cosh b ) \\[2pt]
+ 4 \sinh a_2 \sinh b_2 \cosh^2 a_2 \cosh^2 b_2
\end{array}
\;\right) \tag{Ag} \\[4pt]
= &\phantom{+\;}2 \sinh a_2 \sinh b_2 \sinh c_2 \left(\;
\cosh c - \cosh a\cosh b + ( \cosh a + 1 )( \cosh b + 1 )
\;\right) \tag{Ah} \\[4pt]
= &\phantom{+\;}2 \sinh a_2 \sinh b_2 \sinh c_2 \left(\;
1 + \cosh a + \cosh b + \cosh c
\;\right) \tag{Ai}
\end{align}$$
