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Question Number 178577 by Spillover last updated on 18/Oct/22

$$\mathrm{prove}\mathrm{that}$$$$\left(\mathrm{a}\right)\cosh {}^{-1}\mathrm{x}=\pm \ln (\mathrm{x}+\sqrt{{\mathrm{x}}^2-1})$$$$\left(\mathrm{b}\right)\tanh {}^{-1}\mathrm{x}=\frac{1}{2}\ln \left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right),\mid \mathrm{x}\mid <1$$

Answered by depressiveshrek last updated on 18/Oct/22

$$y=\mathrm{arccosh}x$$$$\cosh y=x$$$$\frac{e^y+e^{-y}}{2}=x$$$$e^y+\frac{1}{e^y}=2x$$$${\left(e^y\right)}^2-2{xe}^y+1=0$$$$e^y=\frac{2x+\sqrt{4x^2-4}}{2}{}^{\ast }$$$$e^y=\frac{2x+2\sqrt{x^2-1}}{2}$$$$e^y=x+\sqrt{x^2-1}$$$$y=\ln (x+\sqrt{x^2-1})$$$$\mathrm{arccosh}x=\ln (x+\sqrt{x^2-1})$$$$$$$$y=\mathrm{arctanh}x$$$$\tanh y=x$$$$\frac{e^y-e^{-y}}{e^y+e^{-y}}=x$$$$e^y-\frac{1}{e^y}={xe}^y+\frac{x}{e^y}$$$${\left(e^y\right)}^2-x{\left(e^y\right)}^2=x+1$$$${\left(e^y\right)}^2(1-x)=x+1$$$${\left(e^y\right)}^2=\frac{x+1}{1-x}$$$$e^y=\sqrt{\frac{x+1}{1-x}}{}^{\ast }$$$$y=\ln \sqrt{\frac{x+1}{1-x}}$$$$\mathrm{arctanh}x=\frac{1}{2}\ln \mid \frac{x+1}{1-x}\mid$$

Commented by Spillover last updated on 18/Oct/22

$$\mathrm{thank}\mathrm{you}$$

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