$\;$ By definition, cosh$x$ = $\frac{e^x + e^{-x}}{2} $ $ \therefore \frac{e^x + e^{-x}}{2} = sec \theta $ or $$ e^x + \frac{1}{e^{x}} = 2 sec \theta $$

$\therefore e^{2x} - (2sec\theta)e^x + 1 = 0 $

$ e^x = \frac{ -(-2sec\theta) \; \;\pm \; \;\sqrt{\;(-2sec\theta)^2 \; \;- \; \;4(1)(1)} }{2} \; \; \ldots [ \because X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \; \; For \; \; a \; \; Quadratic \; \; Equation \; ] $

$ \therefore \; e^x = \frac{2sec\theta \; \; \pm \; \; \sqrt{4(sec^2\theta \; - 1)}}{2} $

= $\frac{2 \;(sec\theta \; \pm \; tan\theta)}{2} \ldots \{ \because \; 1 + tan^2\theta = sec^2\theta\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \& \because tan^2\theta = sec^2\theta - 1 \} $

$ \therefore \; \; e^x = sec\theta \pm tan\theta $

$\; $

By taking log on both sides, we get $x$ = log($ sec\theta \pm tan\theta $) $\;$ However, if $\theta$ exceeds $45^{\circ}$, then value of $sec\theta - tan\theta$ will be negative and taking a log of negative value is impossible. Hence, $x = log( sec\theta + tan\theta )$

Proved.