3.274. $\frac{\sin ^{2} \alpha}{\sin (\alpha-\beta)}+\frac{\sin ^{2} \beta}{\sin (\beta-\alpha)}$.
## Solution.

$$
\frac{\sin ^{2} \alpha}{\sin (\alpha-\beta)}+\frac{\sin ^{2} \beta}{\sin (\beta-\alpha)}=\frac{\sin ^{2} \alpha}{\sin (\alpha-\beta)}-\frac{\sin ^{2} \beta}{\sin (\alpha-\beta)}=
$$

$$
\begin{aligned}
& =\frac{\frac{1-\cos 2 \alpha}{2}-\frac{1-\cos 2 \beta}{2}}{\sin (\alpha-\beta)}=\frac{\cos 2 \beta-\cos 2 \alpha}{2 \sin (\alpha-\beta)}= \\
& =\left[\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]=\frac{-2 \sin (\alpha+\beta) \sin (-(\alpha-\beta))}{2 \sin (\alpha-\beta)}= \\
& =\sin (\alpha+\beta)
\end{aligned}
$$

Answer: $\sin (\alpha+\beta)$.