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
$$\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}}$$

We have
$$
\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}} = 
\frac{\frac{3(\cos^2 \theta + \sin^2 \theta)\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3(\cos^2 \theta + \sin^2 \theta)\cos \theta}{4\cos^3 \theta}} = \\
\frac{\frac{3\cos^2 \theta\sin \theta + 3\sin^3 \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos^3 \theta + 3\sin^2 \theta\cos \theta}{4\cos^3 \theta}} = \\
\frac{\frac{3\cos^2 \theta\sin \theta}{4\cos^3\theta} + \frac{3\sin^3 \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos^3 \theta}{4\cos^3 \theta} -\frac{3\sin^2 \theta\cos \theta}{4\cos^3 \theta}}
$$
and from there, the simplification is straightforward.
