6. Given that $\alpha, \beta$ are acute angles, and
$$
\begin{array}{l}
(1+\sin \alpha-\cos \alpha)(1+\sin \beta-\cos \beta) \\
=2 \sin \alpha \cdot \sin \beta .
\end{array}
$$

Then $\alpha+\beta=$ $\qquad$
6. $\frac{\pi}{2}$.

From the given equation, we have
$$
\begin{array}{l}
\left(2 \sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2} + 2 \sin^2 \frac{\alpha}{2}\right)\left(2 \sin \frac{\beta}{2} \cdot \cos \frac{\beta}{2} + 2 \sin^2 \frac{\beta}{2}\right) \\
\quad=8 \sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2} \cdot \sin \frac{\beta}{2} \cdot \cos \frac{\beta}{2} \\
\Rightarrow\left(\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2}\right)\left(\cos \frac{\beta}{2} + \sin \frac{\beta}{2}\right) \\
\quad=2 \cos \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \\
\Rightarrow\left(1 + \tan \frac{\alpha}{2}\right)\left(1 + \tan \frac{\beta}{2}\right)=2 \\
\Rightarrow \frac{\tan \frac{\alpha}{2} + \tan \frac{\beta}{2}}{1 - \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}}=1 \\
\Rightarrow \tan \frac{\alpha + \beta}{2}=1 \\
\Rightarrow \alpha + \beta=\frac{\pi}{2} .
\end{array}
$$