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Math Help - Trig identity. Probably really easy

  1. #1
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    Trig identity. Probably really easy

    i cant get this

    if i simplify i get opppsites pretty much

    cosx / sinx + 1 = secx - tanx
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  2. #2
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    \frac{\cos{x}}{1+\sin{x}} \cdot \frac{1 - \sin{x}}{1 - \sin{x}} =

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

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

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

    \frac{1}{\cos{x}} - \frac{\sin{x}}{\cos{x}}
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \frac{\cos{x}}{1+\sin{x}} \cdot \frac{1 - \sin{x}}{1 - \sin{x}} =

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

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

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

    \frac{1}{\cos{x}} - \frac{\sin{x}}{\cos{x}}

    I HAD A FEELING IT WOULD BE CONJUGATES TY
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