<s>[INST] 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.
 [/INST] 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.
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