Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 2

Determine the numerical value of a.) $\tan h 0$ and b.) $\tan h 1$

a.) $\tan h 0$

Using Hyperbolic Function


$
\begin{equation}
\begin{aligned}

\tan h x =& \frac{\sin h x}{\cos h x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}
\\
\\
\tan h 0 =& \frac{e^0 - e^{-0}}{e^0 + e^{-0}}
\\
\\
\tan h 0 =& \frac{\displaystyle e^0 - \frac{1}{e^0}}{\displaystyle e^0 + \frac{1}{e^0}}
\\
\\
\tan h 0 =& \frac{1 - 1}{1 + 1}
\\
\\
\tan h 0 =& \frac{0}{2}
\\
\\
\tan h 0 =& 0

\end{aligned}
\end{equation}
$


b.) $\tan h 1$

Using Hyperbolic Function


$
\begin{equation}
\begin{aligned}

\tan h x =& \frac{\sin h x}{\cos h x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}
\\
\\
\tan 1 =& \frac{e^1 - e^{-1}}{e^1 + e^{-1}}
\\
\\
\tan 1 =& \frac{\displaystyle e - \frac{1}{e}}{\displaystyle e + \frac{1}{e}}
\\
\\
\tan 1 =& \frac{\displaystyle \frac{e^2 - 1}{\cancel{e}}}{\displaystyle \frac{e^2 + 1}{\cancel{e}}}
\\
\\
\tan 1 =& \frac{e^2 - 1}{e^2 + 1}
\\
\\
\tan 1 =& 0.7616

\end{aligned}
\end{equation}
$