Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 20

Suppose that $\displaystyle \tan hx = \frac{12}{13}$, find the values of the other hyperbolic functions at $x$.

Solving for $x$

Using Hyperbolic Function


$
\begin{equation}
\begin{aligned}

\tan hx =& \frac{\sin hx}{\cos hx} = \frac{e^x - e^{-x}}{e^x + e^{-x}}
\\
\\
\frac{12}{13} =& \frac{e^x - e^{-x}}{e^x + e^{-x}}
\\
\\
12 (e^x + e^{-x}) =& 13 (e^x - e^{-x})
\\
\\
12 e^x + 12e^{-x} =& 13e^x - 13e^{-x}
\\
\\
25 e^{-x} =& e^x
\\
\\
\frac{25}{e^x} =& e^x
\\
\\
25 =& e^{2x}
\\
\\
\ln 25 =& \ln e^{2x}
\\
\\
\ln (5)^2 =& 2x
\\
\\
2 \ln 5 =& 2x
\\
\\
\frac{\cancel{2} \ln 5}{\cancel{2}} =& \frac{\cancel{2}x}{\cancel{2}}
\\
\\
x =& \ln 5

\end{aligned}
\end{equation}
$



Solving for $\sin hx$


$
\begin{equation}
\begin{aligned}

\sin hx =& \frac{e^x - e^{-x}}{2}
\\
\\
\sin hx =& \frac{e^{\ln 5} - e^{- \ln 5}}{2}
\\
\\
\sin hx =& \frac{5 - 5^{-1}}{2}
\\
\\
\sin hx =& \frac{\displaystyle 5 - \frac{1}{5}}{2}
\\
\\
\sin hx =& \frac{25 - 1}{2 (5)}
\\
\\
\sin hx =& \frac{24}{10}
\\
\\
\sin hx =& \frac{12}{5}

\end{aligned}
\end{equation}
$


Solving for $\cos h x$


$
\begin{equation}
\begin{aligned}

\cos hx =& \frac{e^x + e^{-x}}{2}
\\
\\
\cos hx =& \frac{e^{\ln 5} + e^{- \ln 5}}{2}
\\
\\
\cos hx =& \frac{5 + 5^{-1}}{2}
\\
\\
\cos hx =& \frac{\displaystyle 5 + \frac{1}{5}}{2}
\\
\\
\cos hx =& \frac{25 + 1}{2 (5)}
\\
\\
\cos hx =& \frac{26}{10}
\\
\\
\cos hx =& \frac{13}{5}

\end{aligned}
\end{equation}
$


Solving for $\sec hx$


$
\begin{equation}
\begin{aligned}

\sec hx =& \frac{1}{\cos hx} = \frac{2}{e^x + e^{-x}}
\\
\\
\sec hx =& \frac{2}{e^{\ln 5} + e^{- \ln 5}}
\\
\\
\sec hx =& \frac{2}{5 + 5^{-1}}
\\
\\
\sec hx =& \frac{2}{\displaystyle 5 + \frac{1}{5}}
\\
\\
\sec hx =& \frac{2}{\displaystyle \frac{25 + 1}{5}}
\\
\\
\sec hx =& \frac{2(5)}{26}
\\
\\
\sec hx =& \frac{10}{26}
\\
\\
\sec hx =& \frac{5}{13}

\end{aligned}
\end{equation}
$


Solving for $\csc hx$


$
\begin{equation}
\begin{aligned}

\csc hx =& \frac{1}{\sin hx} = \frac{2}{e^x - e^{-x}}
\\
\\
\csc hx =& \frac{2}{e^{\ln 5} - e^{- \ln 5}}
\\
\\
\csc hx =& \frac{2}{5 - 5^{-1}}
\\
\\
\csc hx =& \frac{2}{\displaystyle 5 - \frac{1}{5}}
\\
\\
\csc hx =& \frac{2}{\displaystyle \frac{25 - 1}{5}}
\\
\\
\csc hx =& \frac{2 (5)}{24}
\\
\\
\csc hx =& \frac{10}{24}
\\
\\
\csc hx =& \frac{5}{12}

\end{aligned}
\end{equation}
$


Solving for $\cot hx$


$
\begin{equation}
\begin{aligned}

\cot hx =& \frac{\cos hx}{\sin hx} = \frac{e^x + e^{-x}}{e^x - e^{-x}}
\\
\\
\cot hx =& \frac{e^{\ln 5} + e^{- \ln 5}}{e^{\ln 5} - e^{- \ln 5}}
\\
\\
\cot hx =& \frac{5 + 5^{-1}}{5 - 5^{-1}}
\\
\\
\cot hx =& \frac{\displaystyle 5 + \frac{1}{5}}{\displaystyle 5 - \frac{1}{5}}
\\
\\
\cot hx =& \frac{\displaystyle \frac{25 + 1}{\cancel{5}}}{\displaystyle \frac
{25 - 1}{\cancel{5}}}
\\
\\
\cot hx =& \frac{26}{24}
\\
\\
\cot hx =& \frac{13}{12}

\end{aligned}
\end{equation}
$