## Rješenje.

Budući da je

$$
\begin{array}{ll}
\operatorname{tg} 192^{\circ}=\operatorname{tg}\left(12^{\circ}+180^{\circ}\right)=\operatorname{tg} 12^{\circ}, & 1 \text { bod } \\
\operatorname{tg} 408^{\circ}=\operatorname{tg}\left(48^{\circ}+360^{\circ}\right)=\operatorname{tg} 48^{\circ}, & 1 \text { bod } \\
\operatorname{tg} 168^{\circ}=\operatorname{tg}\left(-12^{\circ}+180^{\circ}\right)=-\operatorname{tg} 12^{\circ}, & 1 \text { bod }
\end{array}
$$

zadani izraz je jednak

$$
\begin{aligned}
\frac{\operatorname{tg} 12^{\circ}+\operatorname{tg} 48^{\circ}}{1-\operatorname{tg} 12^{\circ} \cdot \operatorname{tg} 48^{\circ}} & =\operatorname{tg}\left(12^{\circ}+48^{\circ}\right) \\
& =\operatorname{tg} 60^{\circ}=\sqrt{3}
\end{aligned}
$$


## Zadatak A-3.1.

Izračunaj

$$
\frac{\operatorname{tg} 192^{\circ}+\operatorname{tg} 48^{\circ}}{1+\operatorname{tg} 168^{\circ} \cdot \operatorname{tg} 408^{\circ}}
$$

