Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 18

Show that $\displaystyle \frac{1 + \tan hx}{1 - \tan hx} = e^{2x}$

Solving for the left-hand side of the equation

Using Hyperbolic Function


$
\begin{equation}
\begin{aligned}

\tan hx =& \frac{\sin hx}{\cos hx} = \frac{e^x - e^{-x}}{e^x + e^{-x}}
\\
\\
\frac{1 + \tan hx}{1 - \tan hx} =& \frac{\displaystyle 1 + \frac{\sin hx}{\cos hx}}{\displaystyle 1 - \frac{\sin hx}{\cos hx}}
\\
\\
\frac{1 + \tan hx}{1 - \tan hx} =& \frac{\displaystyle \frac{\cos hx + \sin hx}{\cancel{\cos hx}}}{\displaystyle \frac{\cos hx - \sin hx}{\cancel{\cos hx}}}
\\
\\
\frac{1 + \tan hx}{1 - \tan hx} =& \frac{\cos hx + \sin hx}{\cos hx - \sin hx}

\end{aligned}
\end{equation}
$


By Hyperbolic Identities


$
\begin{equation}
\begin{aligned}

& \cos hx + \sin hx = e^x \text{ and } \cos hx - \sin hx = e^{-x}
\\
\\
& \frac{1 + \tan hx}{1 - \tan hx} = \frac{e^x}{e^{-x}}
\\
\\
& \frac{1 + \tan hx}{1 - \tan hx} = \frac{e^x}{\displaystyle \frac{1}{e^x}}
\\
\\
& \frac{1 + \tan hx}{1 - \tan hx} = e^x \cdot e^x
\\
\\
& \frac{1 + \tan hx}{1 - \tan hx} = e^{2x}

\end{aligned}
\end{equation}
$