Prove that $\tanh (\cosh^{-1} (x)) = \frac{\sqrt{x^2-1}}{x}$

Question

Prove that

$ \tanh (\cosh^{-1} (x)) = \dfrac{\sqrt{x^2-1}}{x} $ ,

by using that

$ \tanh (x) = \dfrac{e^x-e^{-x}}{e^x+e^{-x}} $ ,

and that

$ \cosh^{-1} (x) = \ln (x+\sqrt{x^2-1}) $ .

After substituting $\cosh^{-1}$ into $x$ in $\tanh (x)$, I'm not sure what to do:

$ \tanh (\cosh^{-1} (x)) = \dfrac{e^{\ln (x+\sqrt{x^2-1})}-e^{-\ln (x+\sqrt{x^2-1})}}{e^{\ln (x+\sqrt{x^2-1})}+e^{-\ln (x+\sqrt{x^2-1})}} = \dfrac{(x+\sqrt{x^2-1})-\dfrac{1}{x+\sqrt{x^2-1}}}{(x+\sqrt{x^2-1})+\dfrac{1}{x+\sqrt{x^2-1}}} $

I've tried multiplying both numerator and denominator with $x+\sqrt{x^2-1}$ and using a binomial theorem of 2nd degree, but I have no idea if that was in the right direction.

Answer 1

There's some tricks to make the algebra tidier, and you really need them for this kind of problem.

Define $s = \sqrt{x^2-1}$ and $c = (x + \sqrt{x^2-1}) = (x + s)$.

Note $s^2 = x^2 - 1$. From here on we use $s$ similar to the way you'd usually use $i$.

Then: \begin{align} \tanh(\cosh^{-1}(x)) &= \tanh(\ln(c)) = \frac{c-\frac 1c}{c + \frac 1c} = \frac{c^2 - 1}{c^2 +1} = \frac{(x+s)^2 -1} {(x+s)^2 +1}\\ &= \frac{x^2 + s^2 -1 + 2xs}{x^2 + s^2 +1 + 2xs}\\ &= \frac{2x^2 - 2 + 2xs}{2 x^2 + 2xs}\\ &= \frac{x(x+s) - 1}{x(x+s)}\\ &= \frac{x(x+s) - 1}{x(x+s)} \cdot \frac{x-s}{x-s}\\ &= \frac{x(x^2-s^2) - (x-s)}{x(x^2-s^2)}\\ &= \frac{x(1) - (x-s)}{x(1)}\\ &= \frac{s}{x}\\ &= \frac{\sqrt{x^2-1}}{x} \end{align}

Answer 2

You have\begin{align}\tanh\left(\cosh^{-1}(x)\right)&=\frac{\sinh\left(\cosh^{-1}(x)\right)}{\cosh\left(\cosh^{-1}(x)\right)}\\&=\frac{\sqrt{\cosh^2\left(\cosh^{-1}(x)\right)-1}}x\\&=\frac{\sqrt{x^2-1}}x.\end{align}