Question
Prove:$$\begin{array}{l}{\text { (a) } \cosh ^{-1} x=\ln (x+\sqrt{x^{2}-1}), \quad x \geq 1} \\ {\text { (b) } \tanh ^{-1} x=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), \quad-1<x<1}\end{array}$$
Step 1
We know that $\cosh y = x$. We can write this in terms of $e$ as $\frac{e^y + e^{-y}}{2} = x$. Show more…
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