[Math] Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$

Question

Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$

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\begin{align}
\tan20^°\tan40^°\tan60^°\tan80^°&=\frac{\sin20^°\sin40^°\sin60^°\sin80^°}{\cos20^°\cos40^°\cos60^°\cos80^°} \\
&=\frac{2^4(\sin20^°\sin40^°\sin60^°\sin80^°)^2}{\sin 160^°}
\end{align}

I am stuck here.

Answer 1

Use $$\tan{(3x)}=\tan{x}\tan{(60^{\circ}+x)}\tan{(60^{\circ}-x)}$$ and let $x=20^\circ$.