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Math Help - Hyperbolic proof

  1. #1
    Junior Member freestar's Avatar
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    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)
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  2. #2
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    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
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  3. #3
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    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!
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