# Thread: Hyperbolic proof

1. ## Hyperbolic proof

$x = ln(sec(\theta) + tan(\theta))$

Show that:
$cosh(x) = sec(\theta)$

I got till here but don't know what to do next.

$cosh(ln(1+sin\theta) - ln(cos\theta)) = sec(\theta)$

2. ## Re: Hyperbolic proof

from given equation
$e^x=\sec(\theta)+\tan(\theta)$ and $e^(-x)=\frac{1}{\sec(\theta)+\tan(\theta)}=\sec(\theta )-\tan(\theta)$
adding them you get required result

3. ## Re: Hyperbolic proof

Hello, freestar!

$x \,=\, \ln(\sec\theta + \tan\theta)$

$\text{Show that: }\:\cosh x \:=\: \sec\theta$

Note that: . $(\sec\theta + \tan\thet)^{-1} \:=\:\frac{1}{\sec\theta + \tan\theta}$

Multiply by $\frac{\sec\theta - \tan\theta}{\sec\theta-\tan\theta}\!:$

. . $\frac{1}{\sec\theta + \tan\theta}\cdot\frac{\sec\theta - \tan\theta}{\sec\theta - \tan\theta} \;=\;\frac{\sec\theta-\tan\theta}{\underbrace{\sec^2\!\theta - \tan^2\!\theta}_{\text{This is 1}}} \;=\;\sec\theta - \tan\theta$

Hence: . $(\sec\theta + \tan\theta)^{-1} \;=\;\sec\theta - \tan\theta$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Use the definition: . $\cosh x \;=\;\frac{e^x + e^{-x}}{2}$

We have: . $\cosh x \:=\:\frac{e^{\ln(\sec\theta+\tan\theta)} + e^{-\ln(\sec\theta+\tan\theta)}}{2}$

. . . . . . . . . . . . . $=\;\frac{e^{\ln(\sec\theta + \tan\theta)} + e^{\ln(\sec\theta + \tan\theta)^{-1}}}{2}$

. . . . . . . . . . . . . $=\;\frac{e^{\ln(\sec\theta + \tan\theta)} + e^{\ln(\sec\theta - \tan\theta)}}{2}$

. . . . . . . . . . . . . $=\;\frac{(\sec\theta + \tan\theta) + (\sec\theta - \tan\theta)}{2}$

. . . . . . . . . . . . . $=\;\frac{2\sec\theta}{2}$

. . . . . . . . . . . . . $=\; \sec\theta$

Edit: Darn, I think I just repeated what Waqarhaider said . . . sorry!