Solution:
In an acute-angled triangle we have $A + B > \frac{\pi}{2}$. Hence we have $\sin A > \sin \left(\frac{\pi}{2} - B\right) = \cos B$ and $\sin B > \cos A$. Using these inequalities we get $(1 - \sin A)(1 - \sin B) < (1 - \cos A)(1 - \cos B)$ and
$$\begin{aligned} \sin A + \sin B &> \cos A + \cos B - \cos A \cos B + \sin A \sin B \\ &= \cos A + \cos B - \cos (A + B) = \cos A + \cos B + \cos C \end{aligned}$$