Question

Prove that the inverse hyperbolic functions can be written in terms of logarithms as shown in Exercises 97–99.

$\tan h^{-1}x=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)$, for $-1<x<1$.

Short answer

We proved$\tan h^{-1}x=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)$, for$-1<x<1$.

Step-by-step solution

Step 1. Given Information 

We have $\tan hy=x$ and we need to prove $t\mathrm{an}{h}^{-1}x=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)$.

Step 2. Proving the statement  

$$\begin{gathered} \tan hy=x \\ \Rightarrow \frac{\sin hy}{\cos hy}=x \\ \Rightarrow \frac{{\displaystyle \frac{e^y-e^{-y}}{2}}}{{\displaystyle \frac{e^y+e^{-y}}{2}}}=x \\ \Rightarrow \frac{e^y-e^{-y}}{e^y+e^{-y}}=x \\ \Rightarrow \frac{e^y-{\displaystyle \frac{1}{e^y}}}{e^y+{\displaystyle \frac{1}{e^y}}}=x \\ \Rightarrow \frac{e^{2y}-1}{e^{2y}+1}=x \\ \Rightarrow e^{2y}-1=x\left(e^{2y}+1\right) \\ \Rightarrow e^{2y}-1=x{e}^{2y}+x \\ \Rightarrow x{e}^{2y}-e^{2y}=-\left(x+1\right) \\ \Rightarrow e^{2y}\left(x-1\right)=-\left(x+1\right) \\ \Rightarrow e^{2y}=\frac{-\left(x+1\right)}{\left(x-1\right)} \\ \Rightarrow e^{2y}=\frac{\left(x+1\right)}{\left(1-x\right)} \\ \Rightarrow \ln \left(e^{2y}\right)=\ln \left(\frac{\left(x+1\right)}{\left(1-x\right)}\right) \\ \Rightarrow 2y=\ln \left(\frac{\left(x+1\right)}{\left(1-x\right)}\right) \\ \Rightarrow y=\frac{1}{2}\ln \left(\frac{\left(x+1\right)}{\left(1-x\right)}\right) \\ \Rightarrow \tan h^{-1}x=\frac{1}{2}\ln \left(\frac{\left(x+1\right)}{\left(1-x\right)}\right) \end{gathered}$$