Exponentielle – Ex3

Exercice 3

Montrer que, pour tout $x\in \R$ on a :$\dfrac{\text{e}^{2x}-1}{\text{e}^{2x}+1} = \dfrac{\text{e}^x-\text{e}^{-x}}{\text{e}^x+\text{e}^{-x}}$

Correction

$\begin{align}
\dfrac{\text{e}^{2x}-1}{\text{e}^{2x}+1} &= \dfrac{\text{e}^x\left(\text{e}^x-\text{e}^{-x}\right)}{\text{e}^x\left(\text{e}^x+\text{e}^{-x}\right)}\\\\
&=\dfrac{\text{e}^x-\text{e}^{-x}}{\text{e}^x+\text{e}^{-x}}
\end{align}$