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Math Help - Differential trig problem

  1. #1
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    Differential trig problem

    Hi, I'm obviously missing something here as I am only getting further awway from the answer.

    show that d/dx (cos x + sin x/ cos x - sin x) = sec^2(pi/4 +x)

    Thanks in advance
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  2. #2
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    Quote Originally Posted by slaypullingcat View Post
    Hi, I'm obviously missing something here as I am only getting further awway from the answer.

    show that d/dx (cos x + sin x/ cos x - sin x) = sec^2(pi/4 +x)

    Thanks in advance
    let's do something with the secant mess first ...

    \sec^2\left(\frac{\pi}{4}+x\right) =

    1 + \tan^2\left(\frac{\pi}{4}+x\right) =

    1 + \left[ \frac{\tan\left(\frac{\pi}{4}\right) + \tan{x}}{1 - \tan\left(\frac{\pi}{4}\right)\tan{x}}\right]^2 =

    1 + \left[\frac{1 + \tan{x}}{1 - \tan{x}}\right]^2 =

    1 + \left[\frac{\cos{x}+\sin{x}}{\cos{x}-\sin{x}}\right]^2 =

    \left[\frac{\cos{x}-\sin{x}}{\cos{x}-\sin{x}}\right]^2+ \left[\frac{\cos{x}+\sin{x}}{\cos{x}-\sin{x}}\right]^2 =

    \frac{(\cos{x}-\sin{x})^2 + (\cos{x}+\sin{x})^2}{(\cos{x}-\sin{x})^2}

    now take the derivative of the left side and see if it matches up.
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  3. #3
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    thank you very much, that was really easy to follow.
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