Prove that $ \tan40° + \sqrt 3 =4 \sin40° $ The equality I'm trying to prove looks like that: $$ \tan40° + \sqrt 3 =4 \sin40° $$
My guess is that $\sqrt3$ can be rewritten as $\tan60°$ and I can use proved in previous exercise formula $$\tan3  \alpha = \frac{(3 - \tan^2\alpha)\tan\alpha}{1 - 3\tan\alpha}$$
But after dealing with it for an hour I feel misled. Some hints would be greatly appreciated.

Let's take $LHS$ as follows  $$LHS=\tan 40^\circ+\sqrt{3}=\tan 40^\circ+\tan 60^\circ$$ $$=\frac{\sin 40^\circ}{\cos 40^\circ}+\frac{\sin 60^\circ}{\cos 60^\circ}$$ $$=\frac{\sin 40^\circ\cos 60^\circ+\cos 40^\circ\sin 60^\circ}{\cos 40^\circ\cos 60^\circ}$$ $$=\frac{\sin (40^\circ+60^\circ)}{\cos 40^\circ\cos 60^\circ}$$ $$=\frac{\sin 2(50^\circ)}{\cos (90^\circ-50^\circ)\left(\frac{1}{2}\right)}$$ $$=\frac{2\times 2\sin 50^\circ \cos 50^\circ}{\sin 50^\circ}$$ $$=4\cos 50^\circ$$$$=\color{blue}{4\sin 40^\circ=RHS}$$
