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Math Help - Proving the identity

  1. #1
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    Proving the identity

    Prove idenity: 1 + sin theta / cos theta = cos theta / 1 - sin theta


    I cant figure out which idenity!?

    Show steps if at all possible thanks in advance.
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  2. #2
    MHF Contributor
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    Hi

    Start with

    cos^2\theta + sin^2\theta = 1

    cos^2\theta = 1 - sin^2\theta

    cos \theta  cos \theta = (1 - sin\theta) (1 + sin\theta)

    \frac{cos \theta}{1 - sin\theta} = \frac{1 + sin\theta}{cos \theta}
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  3. #3
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    ?

    Still a bit confused, The more i think about it, Maybe I just dont understand the goal of proving an idenity, I thought it was to get the problem into the form of one of the basic idenitys such as tan = sin/cos.
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  4. #4
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    Hello, Garface!

    You're expected to know this identity:

    . . \sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \begin{Bmatrix}\sin^2\!\theta \:=\:1 - \cos^2\!\theta \\ \cos^2\!\theta \:=\:1-\sin^2\!\theta \end{Bmatrix}


    Prove the idenity: . \frac{1 + \sin\theta}{\cos\theta} \:=\:\frac{\cos\theta}{1 - \sin\theta }

    Multiply the right side by \frac{1+\sin\theta}{1+\sin\theta}

    . . \frac{\cos\theta}{1-\sin\theta}\cdot\frac{1+\sin\theta}{1 + \sin\theta} \;=\;\frac{\cos\theta(1+\sin\theta)}{1-\sin^2\!\theta} \;=\;\frac{\cos(1+\sin\theta)}{\cos^2\!\theta} \;=\; \frac{1+\sin\theta}{\cos\theta}

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