Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 56

Suppose that $x = \ln (\sec \theta + \tan \theta)$, show that $\sec \theta = \cos hx$

We know that

$\cos hx = \frac{e^x + e^{-x}}{2} $ and $ \sec^2 x - \tan^2 x = 1$

then,


$
\begin{equation}
\begin{aligned}

& (\sec x + \tan x)(\sec x - \tan x) = 1
\\
\\
& \sec x - \tan x = \frac{1}{\sec x + \tan x}

\end{aligned}
\end{equation}
$


Taking the $\ln$ from both sides, we have


$
\begin{equation}
\begin{aligned}

e^x =& e^{\ln (\sec \theta + \tan \theta)}
\\
\\
e^x =& \sec \theta + \tan \theta \qquad \text{ Equation 1}
\\
\\
& \text{ and }
\\
\\
e^{-x} =& \frac{1}{\sec \theta + \tan \theta} \text{ or } e^{-x} = \sec \theta - \tan \theta \qquad \text{ Equation 2}


\end{aligned}
\end{equation}
$


Then, we add Equation 1 and Equation 2, so we get


$
\begin{equation}
\begin{aligned}

e^x + e^{-x} =& \sec \theta + \cancel{\tan \theta} + \sec \theta - \cancel{\tan \theta}
\\
\\
e^x + e^{-x} =& 2 \sec \theta
\\
\\
& \text{ or }
\\
\\
\sec \theta =& \cos hx

\end{aligned}
\end{equation}
$