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

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

$\left.\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}\right]=$

$=\frac{-2 \sin \alpha \sin \beta}{2 \cos \alpha \cos \beta}=-\tan \alpha \cdot \tan \beta$.

Answer: $-\tan \alpha \cdot \tan \beta$.