prove that $\tan2x=\frac{2\sin x\cos x}{1-2\sin^2x}$

Question

Prove that $$\tan2x=\frac{2\sin x\cos x}{1-2\sin^2x}$$

I thought that I could simply write $\sin2x$ divided by $\cos 2x$ gives $\tan 2x$ but the question carries 3 marks. Is there a descriptive proof to solve this?

Answer 1

$2\sin x \cos x=\sin 2x$, while $1-2\sin^2x=\sin^2 x+\cos^2x-2\sin^2x=\cos^2x-\sin^2x=\cos2x$

Thus $$\tan 2x=\frac{\sin 2x}{\cos 2x}$$