# 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}

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