2. Zadatak

Prvo rješenje.

Uredimo desnu stranu dane jednakosti. Prvo ćemo izmnožiti izraze u zagradama i malo pojednostavniti.

$$\begin{aligned} \operatorname{tg}\alpha &=\frac{(1+\operatorname{tg}1^\circ)(1+\operatorname{tg}2^\circ)-2}{(1-\operatorname{tg}1^\circ)(1-\operatorname{tg}2^\circ)-2}\\ &=\frac{1+\operatorname{tg}1^\circ+\operatorname{tg}2^\circ+\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-2}{1-\operatorname{tg}1^\circ-\operatorname{tg}2^\circ+\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-2}\\ &=\frac{\operatorname{tg}1^\circ+\operatorname{tg}2^\circ+\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-1}{-\operatorname{tg}1^\circ-\operatorname{tg}2^\circ+\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-1}. \end{aligned}$$

Nakon sređivanja, podijelimo brojnik i nazivnik s $\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-1$. Slijedi

$$\operatorname{tg}\alpha= \frac{\dfrac{\operatorname{tg}1^\circ+\operatorname{tg}2^\circ}{\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-1}+1} {-\dfrac{\operatorname{tg}1^\circ+\operatorname{tg}2^\circ}{\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-1}+1}.$$

Kako je

$$\operatorname{tg}3^\circ=\frac{\operatorname{tg}1^\circ+\operatorname{tg}2^\circ}{1-\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ},$$

onda je

$$\operatorname{tg}\alpha=\frac{1-\operatorname{tg}3^\circ}{1+\operatorname{tg}3^\circ}.$$

Desnu stranu u danom izrazu, uz činjenicu da je $\operatorname{tg}45^\circ=1$, možemo zapisati kao

$$\operatorname{tg}\alpha=\frac{1-\operatorname{tg}3^\circ}{1+\operatorname{tg}3^\circ}=\frac{\operatorname{tg}45^\circ-\operatorname{tg}3^\circ}{1+\operatorname{tg}45^\circ\cdot\operatorname{tg}3^\circ}=\operatorname{tg}(45^\circ-3^\circ)=\operatorname{tg}42^\circ.$$

Dakle, $\alpha=42^\circ$.

Drugo rješenje.

Uredimo desnu stranu dane jednakosti. Prvo ćemo izmnožiti izraze u zagradama i malo pojednostavniti.

$$\begin{aligned} \operatorname{tg}\alpha &=\frac{(1+\operatorname{tg}1^\circ)(1+\operatorname{tg}2^\circ)-2}{(1-\operatorname{tg}1^\circ)(1-\operatorname{tg}2^\circ)-2}\\ &=\frac{1+\operatorname{tg}1^\circ+\operatorname{tg}2^\circ+\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-2}{1-\operatorname{tg}1^\circ-\operatorname{tg}2^\circ+\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-2}\\ &=\frac{\operatorname{tg}1^\circ+\operatorname{tg}2^\circ+\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-1}{-\operatorname{tg}1^\circ-\operatorname{tg}2^\circ+\operatorname{tg}1^\circ\cdot\operatorname{tg}2^\circ-1}. \end{aligned}$$

Ovu jednakost možemo pisati u obliku

$$\operatorname{tg}\alpha= \frac{\dfrac{\sin1^\circ}{\cos1^\circ}+\dfrac{\sin2^\circ}{\cos2^\circ}+\dfrac{\sin1^\circ}{\cos1^\circ}\cdot\dfrac{\sin2^\circ}{\cos2^\circ}-1} {-\dfrac{\sin1^\circ}{\cos1^\circ}-\dfrac{\sin2^\circ}{\cos2^\circ}+\dfrac{\sin1^\circ}{\cos1^\circ}\cdot\dfrac{\sin2^\circ}{\cos2^\circ}-1}.$$ $$\operatorname{tg}\alpha= \frac{\dfrac{\sin1^\circ\cos2^\circ+\sin2^\circ\cos1^\circ+\sin1^\circ\sin2^\circ-\cos1^\circ\cos2^\circ}{\cos1^\circ\cos2^\circ}} {\dfrac{-\sin1^\circ\cos2^\circ-\sin2^\circ\cos1^\circ+\sin1^\circ\sin2^\circ-\cos1^\circ\cos2^\circ}{\cos1^\circ\cos2^\circ}}.$$ $$\begin{aligned} \operatorname{tg}\alpha &=\frac{(\sin1^\circ\cos2^\circ+\sin2^\circ\cos1^\circ)-(\cos1^\circ\cos2^\circ-\sin1^\circ\sin2^\circ)}{-(\sin1^\circ\cos2^\circ+\sin2^\circ\cos1^\circ)-(\cos1^\circ\cos2^\circ-\sin1^\circ\sin2^\circ)}\\ &=\frac{\sin3^\circ-\cos3^\circ}{-(\sin3^\circ+\cos3^\circ)}\\ &=\frac{\cos3^\circ-\sin3^\circ}{\cos3^\circ+\sin3^\circ}\\ &=\frac{\cos3^\circ(1-\operatorname{tg}3^\circ)}{\cos3^\circ(1+\operatorname{tg}3^\circ)}. \end{aligned}$$

Prema tome je

$$\operatorname{tg}\alpha=\frac{1-\operatorname{tg}3^\circ}{1+\operatorname{tg}3^\circ}.$$

Desnu stranu u danom izrazu, uz činjenicu da je $\operatorname{tg}45^\circ=1$, možemo zapisati kao

$$\operatorname{tg}\alpha=\frac{1-\operatorname{tg}3^\circ}{1+\operatorname{tg}3^\circ}=\frac{\operatorname{tg}45^\circ-\operatorname{tg}3^\circ}{1+\operatorname{tg}45^\circ\cdot\operatorname{tg}3^\circ}=\operatorname{tg}(45^\circ-3^\circ)=\operatorname{tg}42^\circ.$$

Dakle, $\alpha=42^\circ$.