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Question Number 82953 by TawaTawa1 last updated on 26/Feb/20

$$\mathrm{verify}\mathrm{that}:\cosh .{\cosh }^{-1}\left(\mathrm{y}\right)=\mathrm{y},\mathrm{if}\mathrm{y}\in (1,+\mathrm{\infty })$$

Commented by mathmax by abdo last updated on 26/Feb/20

$$wehave{ch}^{-1}\left(y\right)=ln(y+\sqrt{y^2-1})\Rightarrow$$$$ch({ch}^{-1}\left(y\right)=ch(ln(y+\sqrt{y^2-1})$$$$=\frac{e^{ln(y+\sqrt{y^2-1})}+e^{-ln(y+\sqrt{y^2-1})}}{2}=\frac{y+\sqrt{y^2-1+}\frac{1}{y+\sqrt{y^2-1}}}{2}$$$$=\frac{{(y+\sqrt{y^2-1})}^2+1}{2(y+\sqrt{y^2-1})}=\frac{y^2+2y\sqrt{y^2-1}+y^2-1+1}{2(y+\sqrt{y^2-1})}$$$$=\frac{2y^2+2y\sqrt{y^2-1}}{2(y+\sqrt{y^2-1})}=\frac{y^2+y\sqrt{y^2-1}}{y+\sqrt{y^2-1}}=\frac{y(y+\sqrt{y^2-1})}{y+\sqrt{y^2-1}}=y$$

Commented by mathmax by abdo last updated on 26/Feb/20

$$youarewelcome$$

Commented by TawaTawa1 last updated on 26/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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