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.
 A: 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*}
A: You're almost there!
$$\frac{\sec \theta + \tan \theta+\frac{1}{\sec \theta + \tan \theta}{}}{2}=\frac{1}{2}\left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}+\frac{1}{\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}\right)$$ This simplifies to $$\frac{1}{2}\left(\frac{1+\sin\theta}{\cos \theta}+\frac{\cos \theta}{1+\sin\theta}\right)=\frac{1}{2}\left(\frac{1+2\sin \theta+\sin^2\theta+\cos^2\theta}{\cos \theta(1+\sin \theta)}\right)$$ Hence your expression is $$\frac{1}{2}\left(\frac{2(1+\sin\theta)}{\cos \theta(1+\sin \theta)}\right)=\frac{1}{\cos \theta}=\sec \theta$$ as required.
A: Hint:
Why don't we use $$(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$$
$$\implies\dfrac1{\sec\theta+\tan\theta}=?$$
