Geometric method to calculate $\tanh(\cosh^{-1}(x))$

Question

How to calculate $\tanh(\cosh^{-1}(x))$?

And is there a similar method as when calculating: $\tan(\cos^{-1}(x))$? When calculating $\tan(\cos^{-1}(x))$ I would draw a right-triangle like this:

Such that $\tan(\cos^{-1}(x))=\frac{\sqrt{1-x^2}}{x}$

QUESTION: Is there a similar geometric method available to calculate $\tanh(\cosh^{-1}(x))$?

Answer 1

Using the hyperbolic function identity $1+\sinh^2 u = \cosh^2 u$, draw a right triangle with one leg equal to $1$ and the hypoteneuse equal to $x=\cosh u$. Then $\tanh u$ is the ratio of the other leg to the hypoteneuse.