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Math Help - Showing / Proving

  1. #1
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    Showing / Proving

    Show that \frac{d}{dx}(\frac{1+cosx}{sinx})=-\frac{1}{1-cosx}

    I applied the quotient rule but to no avail.. really dont have any idea if my method was right, thanks in advance!
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  2. #2
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    Quote Originally Posted by Punch View Post
    Show that \frac{d}{dx}(\frac{1+cosx}{sinx})=-\frac{1}{1-cosx}

    I applied the quotient rule but to no avail.. really dont have any idea if my method was right, thanks in advance!
    \frac{d}{dx}\left(\frac{1 + \cos{x}}{\sin{x}}\right) = \frac{\sin{x}\,\frac{d}{dx}(1 + \cos{x}) - (1 + \cos{x})\,\frac{d}{dx}(\sin{x})}{\sin^2{x}}

    = \frac{\sin{x}(-\sin{x}) - (1 + \cos{x})\cos{x}}{\sin^2{x}}

     = \frac{-\sin^2{x} - \cos{x} - \cos^2{x}}{\sin^2{x}}

     = \frac{-1 - \cos{x}}{\sin^2{x}}

     = \frac{-(1 + \cos{x})}{1 - \cos^2{x}}

     = \frac{-(1 + \cos{x})}{(1 - \cos{x})(1 + \cos{x})}

     = -\frac{1}{1 - \cos{x}}.
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  3. #3
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    wow didnt know i was halfway there!! thanks
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