6. Given $\alpha, \beta \in \mathbf{R}$, the lines 
$$
\frac{x}{\sin \alpha+\sin \beta}+
\frac{y}{\sin \alpha+\cos \beta}=1 \text { and } \frac{y}{\cos \alpha+\sin \beta}+\frac{y}{\cos \alpha+\cos \beta}=1
$$

intersect at a point on the line $y=-x$, then $\sin \alpha+\cos \alpha+\sin \beta$
$$
+\cos \beta=
$$
6. 0 .

Solution: From the given information, we can assume the intersection point of the two lines to be $\left(x_{0}\right.$, $\left.-x_{0}\right)$, and $\sin \alpha, \cos \alpha$ are the roots of the equation
$$
\frac{x_{0}}{t+\sin \beta}+\frac{-x_{0}}{t+\cos \beta}=1,
$$

which means they are the roots of the equation
$$
\begin{array}{l}
t^{2}+(\cos \beta+\sin \beta) t+\sin \beta \cos \beta-x_{0}(\cos \beta-\sin \beta) \\
=0 \\
\end{array}
$$

Therefore,
$$
\sin \alpha+\cos \alpha=-(\sin \beta+\cos \beta),
$$

which means $\sin \alpha+\cos \alpha+\sin \beta+\cos \beta=0$.