Prove the identity for $\tan3\theta$ 
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$

Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\cos 3 \theta}=\frac{3\sin \theta-4\sin^3\theta}{4\cos^3\theta - 3\cos \theta}$$
To then try and get the whole expression in terms of $\tan\theta$ I multiplied top and bottom of the fraction by $(4\cos^3\theta)^{-1}$. This gave me the following but I'm not now sure how to finish it off
$$\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}}$$

Alternatively, to do it using what you've done so far, rather than applying the double-angle formula twice: 
Note that $\frac{\sin\theta}{\cos^3\theta} = \tan\theta\sec^2\theta = \tan\theta(1 + \tan^2\theta) = \tan\theta + \tan^3\theta$. Alternatively, to do it using what you've done so far, rather than applying the double-angle formula twice: 
note that $\frac{\sin\theta}{\cos^3\theta} = \tan\theta\sec^2\theta = \tan\theta(1 + \tan^2\theta) = \tan\theta + \tan^3\theta$. That will fix up your numerator nicely. Using $\sec^2 \theta = 1 + \tan^2\theta$ directly (since $\frac{\cos\theta}{\cos^3\theta} = \sec^2\theta$) will similarly fix up the numerator, so what you have so far is equal to 
$$\frac{\frac{3}{4}\tan\theta-\frac{1}{4}\tan^3\theta}{\frac{1}{4} - \frac{3}{4}\tan^2\theta} = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta},$$
as required
