6. Given two unequal acute angles $\alpha, \beta$, satisfying
$$
\begin{array}{l}
x \sin \beta + y \cos \alpha = \sin \alpha, \\
x \sin \alpha + y \cos \beta = \sin \beta,
\end{array}
$$

where $\alpha + \beta \neq \frac{\pi}{2}$, and $x, y \in \mathbf{R}$. Then the value of $x^{2} - y^{2}$ is ( ).
(A) -1
(B) 0
(C) 1
(D) does not exist
6.C.

From the problem, we get
$$
\begin{array}{l}
x=\frac{\sin \beta \cdot \cos \alpha-\sin \alpha \cdot \cos \beta}{\sin \alpha \cdot \cos \alpha-\sin \beta \cdot \cos \beta} \\
=\frac{2 \sin (\beta-\alpha)}{\sin 2 \alpha-\sin 2 \beta}=\frac{2 \sin (\beta-\alpha)}{2 \cos (\alpha+\beta) \cdot \sin (\alpha-\beta)} \\
=-\frac{1}{\cos (\alpha+\beta)}=-\sec (\alpha+\beta), \\
y=\frac{\sin ^{2} \alpha-\sin ^{2} \beta}{\sin \alpha \cdot \cos \alpha-\sin \beta \cdot \cos \beta} \\
=\frac{\cos 2 \beta-\cos 2 \alpha}{\sin 2 \alpha-\sin 2 \beta}=\frac{\sin (\alpha+\beta) \cdot \sin (\alpha-\beta)}{\sin (\alpha-\beta) \cdot \cos (\alpha+\beta)} \\
=\tan (\alpha+\beta) .
\end{array}
$$
$$
\text { Therefore, } x^{2}-y^{2}=[-\sec (\alpha+\beta)]^{2}-\tan ^{2}(\alpha+\beta)=1 \text {. }
$$