Completing a proof of a formula for the area of a hyperbolic right triangle Note. The answer that inspired this question had a sign error. That error has been corrected; I'm making the corresponding correction to the formula shown here. (Existing answers acknowledge the error.) --Blue

How can I prove that
  $$
\frac{\tanh a \tanh b \sinh^2 c\sinh a \sinh b \tanh^2c}{\sinh^2 c \tanh^2 c} = \frac{\sinh a \sinh b}{1+\cosh a \cosh b}
$$
  where $\cosh c = \cosh a \cosh b$?

I know $\tanh a = \frac{\sinh a}{\cosh a}$ and $\tanh b = \frac{\sinh b}{\cosh b}$, but where do I go from there?

This relation appears in an answer to this question about the area $K$ of a hyperbolic right triangle with legs $a$, $b$ and hypotenuse $c$. The expression on the right is a target formula for $\sin K$; the expression on the left is an intermediate step at which the answer stops.
 A: It's a bit of a symbolic slog, but let's take things step-by-step.

Note. I've edited this answer to match the corrected version of the question.

$\require{cancel}$
Expressing the tangents in terms of sine and cosine, the denominator on the left becomes
$$\sinh^2c\;\frac{\sinh^2c}{\cosh^2c} = \frac{\sinh^4c}{\cosh^2c} \tag{1}$$
Dividing by that fraction is the same as multiplying by its reciprocal, so we have
$$\frac{\cosh^2c}{\sinh^4c}\left(\frac{\sinh a}{\cosh a}\frac{\sinh b}{\cosh b}\;\sinh^2 c-\sinh a \sinh b \;\frac{\sinh^2c}{\cosh^2 c}\right) \tag{2}$$
Factoring-out $\sinh a\sinh b\sinh^2 c$ (and canceling that last factor appropriately), we have
$$\frac{\sinh a \sinh b \cosh^2 c}{\sinh^2 c}\left(\frac{1}{\cosh a\cosh b}-\frac{1}{\cosh^2c}\right) \tag{3}$$
Using the hyperbolic Pythagorean relation $\cosh c = \cosh a \cosh b$, this is
$$\begin{align}
\frac{\sinh a \sinh b \cosh^2 c}{\sinh^2 c}\left(\frac{1}{\cosh c}-\frac{1}{\cosh^2c}\right)
&=\frac{\sinh a\sinh b\cancel{\cosh^2 c}}{\sinh^2 c}\;\frac{\cosh c-1}{\cancel{\cosh^2 c}} \tag{4} \\[6pt]
&=\frac{\sinh a\sinh b}{\sinh^2 c}\;(\cosh c-1) \tag{5}
\end{align}$$
Now, recall that $\cosh^2 x-\sinh^2x=1$, we can rewrite $\sinh^2 c$:
$$\frac{\sinh a\sinh b (\cosh c-1)}{\cosh^2c-1}=\frac{\sinh a\sinh b \cancel{(\cosh c-1)}}{\cancel{(\cosh c-1)}(\cosh c+1)}=\frac{\sinh a\sinh b}{\cosh c+1} \tag{6}$$
Finally, the Pythagorean relation allows us to rewrite the denominator
$$\frac{\sinh a \sinh b}{1+\cosh a \cosh b} \tag{$\star$}$$
$\square$
A: The original answer you link to has a sign error. The step
$$ \sin [\frac 12 \pi -(\alpha + \beta )] = \sin\alpha\sin\beta-\cos\alpha\cos\beta  $$
is almost correct, except the two sides are additive inverses. You can check this easily. If $\ \alpha = \beta = 0,\ $ then the left side is $\ \sin(\pi/2) = 1\ $ and the right side is
$\ 0 - 1 = -1.\ $ The answer is now corrected.
I have an algebraic method that I have used with good results. Let
$$ \sinh(a) = (A - 1/A)/2,\ \cosh(a) = (A + 1/A)/2\ \textrm{ where } A := e^a.$$
Similarly with $\ b,c.\ $ The equation
 $\ \cosh(c) = \cosh(a) \cosh(b) \ $ when written as a difference is
$$ D := \cosh(c) - \cosh(a) \cosh(b) = 2 A B(1 + C^2) - C\ (1+A^2)(1+B^2). $$
Now the (sign corrected) equation to prove is
$$
\frac{\tanh a \tanh b \sinh^2 c - \sinh a \sinh b \tanh^2c}{\sinh^2 c \tanh^2 c} = \frac{\sinh a \sinh b}{1+\cosh a \cosh b}.
$$
When the right side is subtracted from the left side, the resulting expression is a rational function with $\ D\ $ as a factor in the numerator. Thus the two sides are equal. For this approach it helps to have a computer algebra system that can factor rational functions.
For the curious, the difference is expressed as
$$ \frac{(1-A^2)(1-B^2)((2AB(1+C^2))^2 - (C(1+A^2)(1+B^2))^2)}
   {AB(1+A^2)(1+B^2)(1-C^2)^2(4AB+(1+A^2)(1+B^2))} $$
and $\ D\ $ divides the third difference of two squares factor in the numerator.
