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Prove the identity for $$tan3theta= frac{3tantheta - tan^3 theta}{1-3tan^2 theta}$$

Using de Moivre's theorem I have found that:
$$cos3theta = 4cos^3theta - 3cos theta$$
$$sin 3theta = 3sin theta-4sin^3theta$$

therefore:
$$tan 3theta = frac{sin 3theta}{cos 3 theta}=frac{3sin theta-4sin^3theta}{4cos^3theta - 3cos theta}$$

To then try and get the whole expression in terms of $tantheta$ I multiplied top and bottom of the fraction by $(4cos^3theta)^{-1}$. This gave me the following but I'm not now sure how to finish it off

$$tan3theta =frac{frac{3sin theta}{4cos^3theta} - tan^3theta}{1-frac{3cos theta}{4cos^3 theta}}$$

Alternatively, to do it using what you've done so far, rather than applying the double-angle formula twice:

Note that $frac{sintheta}{cos^3theta} = tanthetasec^2theta = tantheta(1 + tan^2theta) = tantheta + tan^3theta$. Alternatively, to do it using what you've done so far, rather than applying the double-angle formula twice:

note that $frac{sintheta}{cos^3theta} = tanthetasec^2theta = tantheta(1 + tan^2theta) = tantheta + tan^3theta$. That will fix up your numerator nicely. Using $sec^2 theta = 1 + tan^2theta$ directly (since $frac{costheta}{cos^3theta} = sec^2theta$) will similarly fix up the numerator, so what you have so far is equal to

$$frac{frac{3}{4}tantheta-frac{1}{4}tan^3theta}{frac{1}{4} - frac{3}{4}tan^2theta} = frac{3tantheta-tan^3theta}{1-3tan^2theta},$$

as required

Use that $$tan(3x)=frac{tan(x)+tan(2x)}{1-tan(x)tan(2x)}$$ and then
$$tan(x+x)=frac{2tan(x)}{1-tan^2(x)}$$

You have achieved the $cos 3x$ and $sin 3x$ answers by matching the Real and Imaginary components to show them in terms of only $cos x$ and $sin x$ I presume, leading on from an earlier question.

It should be a lot easier if you simply do exactly what you've done for the answers before matching the components

A slightly different approach using de Moivre's theorem would be: notice that $1 + i tan theta = sec theta cdot e^{i theta}$. Therefore, cubing both sides, $(1 + i tan theta)^3 = sec^3 theta cdot e^{3 i theta}$. Now, if you take the "slope" of this complex number, i.e. the imaginary part divided by the real part, from the right hand side you see the result will be $tan(3theta)$. On the other hand, by binomial expansion,
$$(1 + i tan theta)^3 = 1 + 3i tan theta - 3 tan^2 theta - i tan^3 theta = (1 - 3 tan^2 theta) + i (3 tan theta - tan^3 theta).$$

We have
$$
tan3theta =frac{frac{3sin theta}{4cos^3theta} - tan^3theta}{1-frac{3cos theta}{4cos^3 theta}} =
frac{frac{3(cos^2 theta + sin^2 theta)sin theta}{4cos^3theta} - tan^3theta}{1-frac{3(cos^2 theta + sin^2 theta)cos theta}{4cos^3 theta}} = \
frac{frac{3cos^2 thetasin theta + 3sin^3 theta}{4cos^3theta} - tan^3theta}{1-frac{3cos^3 theta + 3sin^2 thetacos theta}{4cos^3 theta}} = \
frac{frac{3cos^2 thetasin theta}{4cos^3theta} + frac{3sin^3 theta}{4cos^3theta} - tan^3theta}{1-frac{3cos^3 theta}{4cos^3 theta} -frac{3sin^2 thetacos theta}{4cos^3 theta}}
$$
and from there, the simplification is straightforward.