8. Zadatak

Prvo rješenje.

$$\begin{aligned} \operatorname{tg}^2\left(\frac{\alpha-\beta}{2}\right) &=\frac{\sin^2\left(\frac{\alpha-\beta}{2}\right)} {\cos^2\left(\frac{\alpha-\beta}{2}\right)}\\ &=\frac{1-\cos(\alpha-\beta)}{1+\cos(\alpha-\beta)}\\ &=\frac{1-(\cos\alpha\cos\beta+\sin\alpha\sin\beta)} {1+(\cos\alpha\cos\beta+\sin\alpha\sin\beta)}\\ &=\frac{1-\frac bc\frac ac-\frac ac\frac bc} {1+\frac bc\frac ac+\frac ac\frac bc}\\ &=\frac{c^2-2ab}{c^2+2ab}\\ &=\frac{a^2+b^2-2ab}{a^2+b^2+2ab}\\ &=\frac{(a-b)^2}{(a+b)^2}. \end{aligned}$$

Iz jednakosti kvadrata i uvjeta $a>b$ slijedi

$$\operatorname{tg}\left(\frac{\alpha-\beta}{2}\right)=\frac{a-b}{a+b}.$$

Drugo rješenje.

Trokut je pravokutan pa je $\alpha+\beta=90^\circ$, tj. $\alpha=90^\circ-\beta$. Stoga

$$\begin{aligned} \operatorname{tg}\left(\frac{\alpha-\beta}{2}\right) &=\operatorname{tg}(45^\circ-\beta)\\ &=\frac{\operatorname{tg}45^\circ-\operatorname{tg}\beta} {1+\operatorname{tg}45^\circ\operatorname{tg}\beta}\\ &=\frac{1-\operatorname{tg}\beta}{1+\operatorname{tg}\beta}. \end{aligned}$$

U pravokutnom trokutu je $\operatorname{tg}\beta=\frac ba$, tj.

$$\operatorname{tg}\left(\frac{\alpha-\beta}{2}\right) =\frac{1-b/a}{1+b/a} =\frac{a-b}{a+b},$$

što je trebalo i dokazati.