Question

funções hiperbólicas mostre que: 1+tgh(x)/1-tgh(x)=e^2x

Answer 1

$t\frac{1+tgh x}{1-tghx}=\frac{ 1+\frac{e^x-e^{-x}}{e^x+e^{-x}} }{ 1-\frac{e^x-e^{-x}}{e^x+e^{-x}} } =\frac{ \frac{e^x+e^{-x}+e^x-e^{-x}}{e^x-e^{-x}} }{ \frac{e^x+e^{-x}-e^x+e^{-x}}{e^x-e^{-x}} } \\ \frac{ \frac{e^x+e^{-x}+e^x-e^{-x}}{e^x+e^{-x}} }{ \frac{e^x+e^{-x}-e^x+e^{-x}}{e^x+e^{-x}} }= \frac{2e^x}{2e^-x} =e^{x+x}=e^{2x} \\ senhx= \frac{e^x-e^{-x}}{2} \\ cosh x=\frac{e^x+e^{-x}}{2}$