Let \(y= \operatorname{csch}(x)\):
\[ y = \frac{2}{e^x - e^{-x}}\]
In order to find the formula for the inverse of \(\operatorname{csch}\) we need to make \(x\) the subject of the above formula.
\[ y(e^x) - y(e^{-x}) - 2 = 0\]
If we multiply both sides by \(e^x\):
\[\begin{gathered} y(e^x)^2 - y - 2e^x = 0 \end{gathered}\]
Let \(v = e^x\):
\[ yv^2 -2v - y = 0 \]
We can use the quadratic formula here:
\[v=\frac{-(-2) ± \sqrt{(-2)^2 - 4(y)(-y)}}{2(y)} = \frac{2 ± \sqrt{4 + 4y^2}}{2y} \]
Simplifying:
\[v = \frac{1}{y} ± \sqrt{\frac{4+4y^2}{4y^2}} = \frac{1}{y} ± \sqrt{\frac{1}{y^2} + 1} \]
Since \(v=e^x\):
\[x = \ln \left( \frac{1}{y} ± \sqrt{\frac{1}{y^2} + 1} \right) \]
Keep in mind the input for \(\ln\) cannot be negative. This means:
\[\operatorname{arcsch(y)} = \ln \left( \frac{1}{y} + \sqrt{\frac{1}{y^2} + 1} \right) \]