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Math Help - another trig id problem

  1. #1
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    another trig id problem

    How do I do this?

    (1+sinx - sin^2x)/(cosx) = cosx + tanx

    This makes no sense....
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  2. #2
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    Re: another trig id problem

    Quote Originally Posted by dazedandmathfused View Post
    How do I do this?

    (1+sinx - sin^2x)/(cosx) = cosx + tanx
    \frac{1+\sin{x}-\sin^2{x}}{\cos{x}} =

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

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

    \frac{\sin{x}}{\cos{x}} + \frac{\cos^2{x}}{\cos{x}} = ?
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  3. #3
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    Re: another trig id problem

    Quote Originally Posted by skeeter View Post
    \frac{1+\sin{x}-\sin^2{x}}{\cos{x}} =

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

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

    \frac{\sin{x}}{\cos{x}} + \frac{\cos^2{x}}{\cos{x}} = ?
    Thank you this identity trig is hard.
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