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Thread: Prove using Trigonometric Identities

  1. #1
    Newbie evanator's Avatar
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    Prove using Trigonometric Identities

    Hi all,

    I am having trouble with the following question. I have filled a few pages with various manipulations and I am no closer to proving the identity. I have obviously missed a trick. Any help would be appreciated.

    First the question:

    Prove that \frac{\cos \theta}{1 - \tan \theta} + \frac{\sin \theta}{1 - \cot \theta} \equiv \sin \theta + \cos \theta

    Let's take the right hand side and see what we can do with it:

    \frac{\sin \theta}{1} + \frac{\cos \theta}{1}

    \frac{\sin^2 \theta}{\sin \theta} + \frac{\cos^2 \theta}{\cos \theta}

    \frac{1 - \cos^2 \theta}{\sin \theta} + \frac{1 - \sin^2 \theta}{\cos \theta}

    \frac{1}{\sin \theta} - \frac{\cos^2 \theta}{\sin \theta} + \frac{1}{\cos \theta} - \frac{\sin^2 \theta}{\cos \theta}

    \frac{\sin \theta - \sin \theta \cos^2 \theta}{\sin^2 \theta} + \frac{\cos \theta - \cos \theta \sin^2 \theta}{\cos^2 \theta}

    \frac{\sin \theta -\sin \theta \cos^2 \theta}{1-\cos^2 \theta} + \frac{\cos \theta - \cos \theta \sin^2 \theta}{1 - \sin^2 \theta}

    I can't really see where to go with it.

    Regards,

    Evanator
    Last edited by Jhevon; Oct 24th 2010 at 09:15 AM.
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by evanator View Post
    Hi all,

    I am having trouble with the following question. I have filled a few pages with various manipulations and I am no closer to proving the identity. I have obviously missed a trick. Any help would be appreciated.

    First the question:

    Prove that \frac{\cos \theta}{1 - \tan \theta} + \frac{\sin \theta}{1 - \cot \theta} \equiv \sin \theta + \cos \theta
    \displaystyle \frac{\cos{t}}{1-\tan{t}} \cdot \frac{\cos{t}}{\cos{t}} + \frac{\sin{t}}{1-\cot{t}} \cdot \frac{\sin{t}}{\sin{t}} =

    \displaystyle \frac{\cos^2{t}}{\cos{t} - \sin{t}} + \frac{\sin^2{t}}{\sin{t} - \cos{t}} =

    \displaystyle \frac{\cos^2{t} - \sin^2{t}}{\cos{t} - \sin{t}} =

    last two steps I leave for you ...
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  3. #3
    MHF Contributor Unknown008's Avatar
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    It is usually much more easier to go from 'big ones' to 'smaller ones', that is the one on the left.

    \begin{array}{lcl}<br />
\dfrac{\cos \theta}{1 - \tan \theta} + \dfrac{\sin \theta}{1 - \cot \theta} &=& \dfrac{\cos \theta}{1 - \frac{\sin\theta}{\cos\theta}} + \dfrac{\sin \theta}{1 - \frac{\cos\theta}{\sin\theta}} \\<br />
&& \\<br />
&=& \dfrac{\cos \theta}{ \frac{\cos\theta - \sin\theta}{\cos\theta}} + \dfrac{\sin \theta}{\frac{sin\theta - \cos\theta}{\sin\theta}} \\<br />
&& \\<br />
&=& \dfrac{\cos^2 \theta}{\cos\theta - \sin\theta} + \dfrac{\sin^2 \theta}{\sin\theta - \cos\theta} \\<br />
&& \\<br />
&=& \dfrac{\cos^2 \theta}{\cos\theta - \sin\theta} - \dfrac{\sin^2 \theta}{\cos\theta - \sin\theta} \\<br />
&& \\<br />
&=& \dfrac{\cos^2 \theta - \sin^2\theta}{\cos\theta - \sin\theta}\end{array}<br />
\end{array}

    Can you complete it now?

    EDIT: A tad too late
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  4. #4
    Newbie evanator's Avatar
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    \frac{\cos^2 \theta - \sin^2 \theta}{\cos \theta - \sin \theta} =

    \frac{(\cos \theta + \sin \theta)(\cos \theta - \sin \theta)}{\cos \theta - \sin \theta} =

    \cos \theta + \sin \theta

    Thank you both, and thank you to whoever moved my thread. I'm always quite annoyed when I see how straightforward it is, but I'm sure I'll get a feel for it with time.
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