Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$ Prove that $\tan(6^\circ)\tan(42^\circ) = \tan(12^\circ) \tan(24^\circ)$. 
I don't know how to approach this problem. One approach might be to note that $42-6= 24+12$, and then apply the identities for $\tan(A+B)$ and $\tan(A-B)$, but it just makes it more complex:
$$\frac{\tan(42^{\circ}) - \tan(6^{\circ})}{1+\tan(42^{\circ})\tan(6^{\circ})}= \frac{\tan(24^{\circ}) + \tan(12^{\circ})}{1 - \tan(24^{\circ})\tan(12^{\circ})}.$$
Can anyone here please provide a hint? 

Here is a proof that I got after back calculation.
Let $\theta=36^{\circ}$ Then, one can show that $\theta$ satisfies the following equation $$(\cos \theta +1)(4\cos^2\theta -2\cos\theta-1)=0$$since $\cos 36^{\circ}>0$, $\theta $ satisfies $$4\cos^2\theta-2\cos \theta-1=0\\ \Rightarrow 2\cos 2\theta-2\cos \theta=-1\ \\ \Rightarrow \cos 2\theta-\cos \theta=-\frac{1}{2}$$ Let $\phi=12^{\circ}$. Then $\theta=3\phi$ and $$\cos 6\phi-\cos 3\phi=-\frac{1}{2}\\ \Rightarrow \cos 6\phi+1=\cos 3\phi+\frac{1}{2}=\cos 3\phi+\cos 5\phi\quad (\because 5\phi=60^{\circ})\\ \Rightarrow 2\cos^2 3\phi=2\cos 4\phi \cos \phi\\ \Rightarrow \frac{\cos 36^{\circ}}{\cos 48^{\circ}}=\frac{\cos 12^{\circ}}{\cos 36^{\circ}}\\ \Rightarrow \frac{\cos 6^{\circ}\cos 42^{\circ}+\sin 6^{\circ}\sin 42^{\circ}}{\cos 6^{\circ}\cos 42^{\circ}-\sin 6^{\circ}\sin 42^{\circ}}=\frac{\cos 24^{\circ}\cos 12^{\circ}+\sin 24^{\circ}\sin 12^{\circ}}{\cos 24^{\circ}\cos 12^{\circ}-\sin 24^{\circ}\sin 12^{\circ}}\\ \Rightarrow \frac{\cos 6^{\circ}\cos 42^{\circ}}{\sin 6^{\circ}\sin 42^{\circ}}=\frac{\cos 24^{\circ}\cos 12^{\circ}}{\sin 24^{\circ}\sin 12^{\circ}}\quad (\mbox{by Componendo-Dividendo})\\ \Rightarrow \tan 6^{\circ}\tan 42^{\circ}=\tan 24^{\circ}\tan 12^{\circ}\hspace{6cm}\Box$$
