1. Zadatak

Prvo rješenje.

Kako je $\operatorname{ctg}62^\circ=\operatorname{tg}28^\circ$

$$\frac{\operatorname{tg}58^\circ-\operatorname{tg}28^\circ}{1+\operatorname{tg}58^\circ\operatorname{ctg}62^\circ}=\frac{\operatorname{tg}58^\circ-\operatorname{tg}28^\circ}{1+\operatorname{tg}58^\circ\cdot\operatorname{tg}28^\circ}=\operatorname{tg}(58^\circ-28^\circ)=\operatorname{tg}30^\circ=\frac{\sqrt3}3.$$

Drugo rješenje.

$$\begin{aligned} \frac{\operatorname{tg}58^\circ-\operatorname{tg}28^\circ}{1+\operatorname{tg}58^\circ\operatorname{ctg}62^\circ} &=\frac{\frac{\sin58^\circ}{\cos58^\circ}-\frac{\sin28^\circ}{\cos28^\circ}}{1+\frac{\sin58^\circ\cos62^\circ}{\cos58^\circ\sin62^\circ}}=\frac{\frac{\sin(58^\circ-28^\circ)}{\cos58^\circ\cos28^\circ}}{\frac{\sin(58^\circ+62^\circ)}{\cos58^\circ\sin62^\circ}}=\\ &=\frac{\sin30^\circ\sin62^\circ}{\sin120^\circ\cos28^\circ}=\frac{\frac12}{\frac{\sqrt3}2}=\frac{\sqrt3}3. \end{aligned}$$