Thread: Rearranging fractions to isolate function

1. 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.

2. Re: Rearranging fractions to isolate function

Originally Posted by NitrousUK
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. 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!

4. Re: Rearranging fractions to isolate function

Originally Posted by NitrousUK
$\tan^2\theta$
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