11.9. Simplify the expression

$$
\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}+\sqrt{3} \operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}
$$
11.9. Let's show that $\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}+\sqrt{3} \operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}=\sqrt{3}$, i.e., $\sqrt{3}=\frac{\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}}{1-\operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}}$. Indeed, $\frac{\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}}{1-\operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}}=\operatorname{tg} 60^{\circ}=\sqrt{3}$.

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