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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.

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    $\begingroup$ So far so good. Get a common denominator and write this as a single fraction. $\endgroup$ – Matthew Leingang Nov 10 '17 at 20:17
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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*}

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  • $\begingroup$ this is identically with my way and idea $\endgroup$ – Dr. Sonnhard Graubner Nov 10 '17 at 20:25
  • $\begingroup$ @Dr.SonnhardGraubner Yes, you're right that we had the same idea. I just showed a few more steps. I didn't see your answer before I posted mine, nor did I downvote your answer before you deleted it. $\endgroup$ – Matthew Leingang Nov 10 '17 at 20:27
  • $\begingroup$ No, @DrS You wrote the second line only, with now explanation. That's not a very difficult feat to accomplish. You do not own rights to an approach, when others independently use it too. $\endgroup$ – amWhy Nov 10 '17 at 20:27
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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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  • $\begingroup$ using \theta instead of \Theta is far more standard, and makes the answer more readable (when I read it, e.g., I was very distracted at how large (and also more bold, $\Theta$ is compared to $\theta.$ Otherwise, nice answer. $\endgroup$ – amWhy Nov 10 '17 at 20:23
  • $\begingroup$ OP previously used $\Theta$ instead of $\theta$. I will edit it now. $\endgroup$ – TheSimpliFire Nov 10 '17 at 20:24
  • $\begingroup$ Ahh, I didn't see the original question! $\endgroup$ – amWhy Nov 10 '17 at 20:25
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Hint:

Why don't we use $$(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$$

$$\implies\dfrac1{\sec\theta+\tan\theta}=?$$

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  • $\begingroup$ genius! simplest answer by far (+1) $\endgroup$ – TheSimpliFire Dec 22 '17 at 7:50

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