Answer
$sec~\theta = cosh~x$
Work Step by Step
Suppose that $x = ln(sec~\theta+tan~\theta)$
Then:
$e^x = sec~\theta+tan~\theta$
$e^x = \frac{1}{cos~\theta}+\frac{sin~\theta}{cos~\theta}$
$e^x = \frac{1+sin~\theta}{cos~\theta}$
$e^x~cos~\theta-1 = sin~\theta$
$e^x~cos~\theta-1 = \sqrt{1-cos^2~\theta}$
$e^{2x}~cos^2~\theta-2e^x~cos~\theta+1 = 1-cos^2~\theta$
$(e^{2x}+1)~cos^2~\theta = 2e^x~cos~\theta$
$(e^{2x}+1)~cos~\theta = 2e^x$
$cos~\theta = \frac{2e^x}{e^{2x}+1}$
$sec~\theta = \frac{e^{2x}+1}{2e^x}$
$sec~\theta = \frac{e^x+1/e^x}{2}$
$sec~\theta = \frac{e^x+e^{-x}}{2}$
$sec~\theta = cosh~x$
