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Math Help - Rearranging fractions to isolate function

  1. #1
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    Rearranging fractions to isolate function

    I'm working through past papers for A-Level C3 maths and i'm stuck on how the answer book arrived at the following rearrangement.

    \frac{\tan^2\theta-3}{1-3\tan^2\theta}=k^2

    Rearranged into form:
    \tan^2\theta=\frac{k^2+3}{1+3k^2}

    I'm sure there's some multiply divide by terms but I can't figure out which. Any help would be massively appreciated. Thanks.
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  2. #2
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    Re: Rearranging fractions to isolate function

    Quote Originally Posted by NitrousUK View Post
    I'm working through past papers for A-Level C3 maths and i'm stuck on how the answer book arrived at the following rearrangement.

    \frac{\tan^2\theta-3}{1-3\tan^2\theta}=k^2

    Rearranged into form:
    \tan^2\theta=\frac{k^2+3}{1+3k^2}

    I'm sure there's some multiply divide by terms but I can't figure out which. Any help would be massively appreciated. Thanks.
    \frac{\tan^2\theta-3}{1-3\tan^2\theta}=k^2

    \tan^2\theta-3 = (1-3\tan^2\theta)k^2

    \tan^2\theta-3 = k^2-3k^2\tan^2\theta

    \tan^2\theta + 3k^2\tan^2\theta = k^2+3

    \tan^2\theta(1+3k^2) = k^2+3

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  3. #3
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    Re: Rearranging fractions to isolate function

    Factor the tan! Of course. I was stuck in thinking I was to reduce the occurence of tan by cancelling in a fraction. Thank you!
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    Re: Rearranging fractions to isolate function

    Quote Originally Posted by NitrousUK View Post
    \tan^2\theta
    Well, you'd find it easier if you let x = above; then:

    (x^2 - 3) / (1 - 3x^2) = k^2

    If not easier, then less confusing
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