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Math Help - Proving identities 2

  1. #1
    Member Furyan's Avatar
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    Proving identities 2

    Hello, me again

    I'm stuck on another identity:

    \dfrac{1 - \sin\theta}{\cos\theta} \equiv \dfrac{1}{\sec\theta + \tan\theta}

    I got:

    LHS \equiv \dfrac{1}{\cos\theta} - \dfrac{\sin\theta}{\cos\theta} \equiv \sec\theta - \tan\theta

    I tried starting with RHS and couldn’t do it that way either. I'd really appreciate some help.


    Thank you
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  2. #2
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    Re: Proving identities 2

    Quote Originally Posted by Furyan View Post
    \dfrac{1 - \sin\theta}{\cos\theta} \equiv \dfrac{1}{\sec\theta + \tan(\theta)}
    Change the RHS to \frac{\cos(\theta)}{1+\sin(\theta)}
    Then multiply by \frac{1-\sin(\theta)}{1-\sin(\theta)}.
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  3. #3
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    Re: Proving identities 2

    Quote Originally Posted by Furyan View Post
    Hello, me again

    I'm stuck on another identity:

    \dfrac{1 - \sin\theta}{\cos\theta} \equiv \dfrac{1}{\sec\theta + \tan\theta}

    I got:

    LHS \equiv \dfrac{1}{\cos\theta} - \dfrac{\sin\theta}{\cos\theta} \equiv \sec\theta - \tan\theta

    I tried starting with RHS and couldn’t do it that way either. I'd really appreciate some help.


    Thank you
    That works. Next multiply numerator and denominator by \sec \theta + \tan \theta to give \dfrac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\sec \theta + \tan \theta}

    You can simplify that numerator using the difference of two squares (this is a nifty little formula!) which in turn can have an identity used on it - try rearranging \tan^2 \theta + 1 = \sec^2 \theta if you get stuck
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    Member Furyan's Avatar
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    Re: Proving identities 2

    Quote Originally Posted by Plato View Post
    Change the RHS to \frac{\cos(\theta)}{1+\sin(\theta)}
    Then multiply by \frac{1-\sin(\theta)}{1-\sin(\theta)}.
    Thank you Plato,

    That worked like a charm, really amazing. Can I ask did you choose \dfrac{1 - \sin\theta}{1 - \sin\theta} because you could see it would give you \cos^2\theta in the denominator? I haven't used a manipulation like that in any other question I've done, but I have a feeling if I had they may have been very much easier. I have a similar one I was stuck on I'll try it on that and see if it helps.

    Thanks again.
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    Re: Proving identities 2

    Quote Originally Posted by Furyan View Post
    Can I ask did you choose \dfrac{1 - \sin\theta}{1 - \sin\theta} because you could see it would give you \cos^2\theta in the denominator?
    To be truthful, over fifty years ago I won a citywide identities proving contest. I just never lost the ability to see the needed trick.
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  6. #6
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    Re: Proving identities 2

    Quote Originally Posted by e^(i*pi) View Post
    That works. Next multiply numerator and denominator by \sec \theta + \tan \theta to give \dfrac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\sec \theta + \tan \theta}

    You can simplify that numerator using the difference of two squares (this is a nifty little formula!) which in turn can have an identity used on it - try rearranging \tan^2 \theta + 1 = \sec^2 \theta if you get stuck
    Thank you e^i*pi, that worked like a charm too. Thanks for the hint about \sec^2\theta - \tan^2\theta \equiv 1 I hadn't used that and might not have seen it. Now I know I should try using that equivalent fraction technique if I'm really stuck, which I have been on another on for some time. I hope it works on it. The difference of two squares has come up a lot. Plato's method above uses it in the denominator.

    Thanks again
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  7. #7
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    Re: Proving identities 2

    Quote Originally Posted by Plato View Post
    To be truthful, over fifty years ago I won a citywide identities proving contest. I just never lost the ability to see the needed trick.
    Thanks, that explains how you can solve in a micro second something I've been working on for hours. I was working on others that I could do in that time too though.
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