Problem 1148

Proof 1: The given equation can be transformed into
$$\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \operatorname{tg} \beta}=\frac{\frac{2 \operatorname{tg} \beta}{1+\operatorname{tg}^{2} \beta}}{5-\frac{1-\operatorname{tg}^{2} \beta}{1+\operatorname{tg}^{2} \beta}} .$$

Let $\operatorname{tg} \alpha=m, \operatorname{tg} \beta=n$, then the above equation can be transformed into $\frac{m-n}{1+m n}=\frac{2 n}{4+6 n^{2}}$, i.e., $(3 n-2 m)\left(n^{2}+1\right)=0$, hence $3 n=2 m$, i.e., $2 \operatorname{tg} \alpha=3 \operatorname{tg} \beta$.
Proof 2: $\operatorname{tg} \alpha=\operatorname{tg}(\alpha-\beta+\beta)=\frac{\operatorname{tg}(\alpha-\beta)+\operatorname{tg} \beta}{1-\operatorname{tg}(\alpha-\beta) \operatorname{tg} \beta}$ $=\frac{6(1-\cos 2 \beta)}{4 \sin 2 \beta}=\frac{3}{2} \operatorname{tg} \beta$.

Therefore, the equation holds.