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.
Best Answer
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 $$\frac{\sinh a \sinh b}{1+\cosh a \cosh b} \tag{$\star$}$$
$\square$