Prove that $\cosh(x)=\sec(\theta )$ if $x=\ln(\sec \theta + \tan \theta)$ I'm trying to prove that $\cosh(x)=\sec(\theta )$ if $x=x=\ln(\sec \theta + \tan \theta)$. I've substituted the value of $x$ into $\cosh(x)$ to get $$\frac{e^{\ln(\sec \theta + \tan \theta)}+e^{-\ln(\sec \theta + \tan \theta)}}{2}$$ and simplified to get $$\frac{(\sec \theta +\tan \theta)+\frac{1}{(\sec \theta + \tan \theta)}}{2}$$. However, I do not know where to continue from here. Help would be greatly appreciated.

You got to
$$
    \cosh x = \frac{1}{2}\left(\sec\theta + \tan\theta + \frac{1}{\sec\theta+\tan\theta}\right)
$$
Write this is as single fraction:
$$
\frac{1}{2}\left(\sec\theta + \tan\theta + \frac{1}{\sec\theta+\tan\theta}\right)
= \frac{(\sec\theta + \tan\theta)^2 + 1}{2(\sec\theta + \tan\theta)}
$$
Expand and remember that $1 + \tan^2 \theta = \sec^2\theta$:
\begin{align*}
\frac{(\sec\theta + \tan\theta)^2 + 1}{2(\sec\theta + \tan\theta)}
&= \frac{\sec^2\theta + 2\tan\theta\sec\theta + \tan^2\theta + 1}{2(\sec\theta + \tan\theta)}
\\&= \frac{2\sec^2\theta + 2 \sec\theta\tan\theta}{2(\sec\theta + \tan\theta)}
\\&= \frac{(2\sec\theta)(\sec\theta + \tan\theta)}{2(\sec\theta + \tan\theta)}
   = \sec\theta
\end{align*}
