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Math Help - pythagorean identity

  1. #1
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    pythagorean identity

    I have an example that is about proving

    \frac{sin \theta}{cot \theta cos^2 \theta - cot \theta} - sec \theta} = 0

    it factors the denominator of the left term into \frac{sin \theta}{cot \theta ( cos^2 \theta - 1)}

    then says replace (cos^2 \theta - 1) with sin^2 \theta

    Is this correct?

    wouldn't cos^2 \theta - 1 = -sin^2 \theta ??
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  2. #2
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    Re: pythagorean identity

    Quote Originally Posted by Jonroberts74 View Post
    I have an example that is about proving

    \frac{sin \theta}{cot \theta cos^2 \theta - cot \theta} - sec \theta} = 0

    it factors the denominator of the left term into \frac{sin \theta}{cot \theta ( cos^2 \theta - 1)}

    then says replace (cos^2 \theta - 1) with sin^2 \theta

    Is this correct?

    wouldn't cos^2 \theta - 1 = -sin^2 \theta ??
    you are correct.
    Thanks from Jonroberts74
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  3. #3
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    Re: pythagorean identity

    Hello, Jonroberts74!

    The given equation is not true.


    \text{Prove: }\:\frac{\sin\theta}{\cot\theta\cos^2\!\theta -\cot\theta} + \sec\theta} \:=\: 0
    . . . . . . . . . . . . . . . . . . . . . {\color{red}\uparrow}

    We have: . \frac{\sin\theta}{\cot\theta(\cos^2\theta-1)} + \sec\theta \;=\;\frac{\sin\theta}{\cot\theta(\text{-}\sin^2\theta)} + \sec\theta \;=\;\dfrac{\sin\theta}{\dfrac{\cos\theta}{\sin \theta}(\text{-}\sin^2\theta)}+ \sec\theta

    . . . . . =\; \dfrac{\sin\theta}{\text{-}\cos\theta\sin\theta}+\sec\theta \;=\;\frac{\text{-}1}{\cos\theta} + \sec\theta \;=\; \text{-}\sec\theta + \sec\theta \;=\;0
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  4. #4
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    Re: pythagorean identity

    so many books with errors.
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