$\sin (\alpha+\beta) \sin (\alpha-\beta)-(\sin \alpha+\sin \beta)(\sin \alpha-\sin \beta)=?$
$$
\begin{gathered}
\sin (\alpha+\beta) \sin (\alpha-\beta)=(\sin \alpha \cos \beta+\cos \alpha \sin \beta)(\sin \alpha \cos \beta-\cos \alpha \sin \beta)= \\
=\sin ^{2} \alpha \cos ^{2} \beta-\cos ^{2} \alpha \sin ^{2} \beta=\sin ^{2} \alpha\left(1-\sin ^{2} \beta\right)-\left(1-\sin ^{2} \alpha\right) \sin ^{2} \beta= \\
=\sin ^{2} \alpha-\sin ^{2} \beta
\end{gathered}
$$

therefore

$$
\begin{gathered}
\sin (\alpha+\beta) \sin (\alpha-\beta)-(\sin \alpha+\sin \beta)(\sin \alpha-\sin \beta)= \\
=\sin ^{2} \alpha-\sin ^{2} \beta-\left(\sin ^{2} \alpha-\sin ^{2} \beta\right)=0
\end{gathered}
$$

(Schudich Lajos, Eperjes.)