1. Zadatak

a) Primjenjujemo adicijski teorem za tangens: $$\begin{aligned} \operatorname{tg}3x&=\operatorname{tg}(2x+x)=\frac{\operatorname{tg}2x+\operatorname{tg}x}{1-\operatorname{tg}2x\cdot\operatorname{tg}x}\\ &=\frac{\frac{2\operatorname{tg}x}{1-\operatorname{tg}^2x}+\operatorname{tg}x}{1-\operatorname{tg}x\cdot\frac{2\operatorname{tg}x}{1-\operatorname{tg}^2x}}\\ &=\frac{\frac{3\operatorname{tg}x-\operatorname{tg}^3x}{1-\operatorname{tg}^2x}}{\frac{1-\operatorname{tg}^2x-2\operatorname{tg}^2x}{1-\operatorname{tg}^2x}}=\frac{3\operatorname{tg}x-\operatorname{tg}^3x}{1-3\operatorname{tg}^2x} \end{aligned}$$ S druge strane, vrijedi: $$\begin{aligned} &\operatorname{tg}x\cdot\operatorname{tg}(60^\circ+x)\cdot\operatorname{tg}(60^\circ-x)\\ &=\operatorname{tg}x\cdot\frac{\operatorname{tg}60^\circ+\operatorname{tg}x}{1-\operatorname{tg}60^\circ\cdot\operatorname{tg}x}\cdot\frac{\operatorname{tg}60^\circ-\operatorname{tg}x}{1+\operatorname{tg}60^\circ\operatorname{tg}x}\\ &=\operatorname{tg}x\cdot\frac{\sqrt3+\operatorname{tg}x}{1-\sqrt3\operatorname{tg}x}\cdot\frac{\sqrt3-\operatorname{tg}x}{1+\sqrt3\cdot\operatorname{tg}x}\\ &=\operatorname{tg}x\cdot\frac{3-\operatorname{tg}^2x}{1-3\operatorname{tg}^2x}=\frac{3\operatorname{tg}x-\operatorname{tg}^3x}{1-3\operatorname{tg}^2x}. \end{aligned}$$ Iz jednakosti desnih strana slijedi istinitost zadane tvrdnje.

b) Ako u jednakost dokazanu pod a) stavimo $x=20^\circ$, onda dobijemo: $$\begin{aligned} \operatorname{tg}60^\circ&=\operatorname{tg}20^\circ\cdot\operatorname{tg}80^\circ\cdot\operatorname{tg}40^\circ\ /\cdot\operatorname{tg}60^\circ\\ \operatorname{tg}^260^\circ&=\operatorname{tg}20^\circ\cdot\operatorname{tg}40^\circ\cdot\operatorname{tg}60^\circ\cdot\operatorname{tg}80^\circ\\ 3&=\operatorname{tg}20^\circ\cdot\operatorname{tg}40^\circ\cdot\operatorname{tg}60^\circ\cdot\operatorname{tg}80^\circ. \end{aligned}$$