Is it true that $\sinh^{-1}\Big(\frac{z}{\sqrt{1-z^2}}\Big)=\tanh^{-1}(z)$?

Question

How does one prove that it is?

Answer 1

Let $\tanh ^{-1}z=y\implies \tanh y=z$

So, $$\frac{\cosh y}1=\frac{\sinh y} z=\frac{\sqrt{\cosh^2 y-\sinh ^2y}}{\sqrt{1-z^2}}=\frac 1{\sqrt{1-z^2}}$$ as $\cosh^2 y-\sinh ^2y=1$

So, $y=\cosh ^{-1}\frac1{\sqrt{1-z^2}}=\sinh ^{-1}\frac z{\sqrt{1-z^2}}$

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Or, $$\frac z{\sqrt{1-z^2}}=\frac {\tanh y}{\sqrt{1-\tanh^2 y}}=\frac{\sinh y}{\cosh y\sqrt{1-\frac{\sinh^2 y}{\cosh^2y}}}=\frac{\sinh y}{\sqrt{\cosh^2 y-\sinh ^2y}}=\sinh y$$