Completing a proof of a formula for the area of a hyperbolic right triangle

Question

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?

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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.

Answer 1

It's a bit of a symbolic slog, but let's take things step-by-step.

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

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$\require{cancel}$ 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 $$\frac{\sinh a \sinh b}{1+\cosh a \cosh b} \tag{$\star$}$$

$\square$

Answer 2

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.