3.303. $1-\sin ^{2} \alpha-\sin ^{2} \beta+2 \sin \alpha \sin \beta \cos (\alpha-\beta)$.
## Solution.

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

Answer: $\cos ^{2}(\alpha-\beta)$.