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  1. #1
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    Prove

    Prove  <br />
sin2A-tanA=cos2AtanA<br />
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  2. #2
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    Quote Originally Posted by Punch View Post
    Prove  <br />
sin2A-tanA=cos2AtanA<br />
    \sin{2x} - \tan{x} = 2\sin{x}\cos{x} - \frac{\sin{x}}{\cos{x}}

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

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

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

     = \cos{2x}\tan{x}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    \sin{2x} - \tan{x} = 2\sin{x}\cos{x} - \frac{\sin{x}}{\cos{x}}

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

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

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

     = \cos{2x}\tan{x}.
    Thanks amended a minor error
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