As we need to eliminate $\beta,$ write $\tan\beta=\tan\{\alpha-(\alpha-\beta)\}$ and expand.
For the RHS, $$\frac{n\sin\alpha\cos\alpha}{1-n\sin^2\alpha}=\dfrac{\dfrac{n\sin\alpha\cos\alpha}{\cos^2\alpha}}{\dfrac{1-n\sin^2\alpha}{\cos^2\alpha}}=\dfrac{n\tan^2\alpha}{1+(1-n)\tan^2\alpha}$$
Best Answer
It's always the case that $\tan2x=2\tan x/(1-\tan^2x)$. If $\tan^2x =2\tan x+1$, then $2\tan x=\tan^2x-1$ and hence
$$\tan2x={2\tan x\over 1-\tan^2x}={\tan^2x-1\over1-\tan^2x}=-1$$