1. The maximum value of the function $y=\frac{\sin \alpha \cdot \cos \alpha}{1+\cos \alpha}\left(\alpha \in\left(0, \frac{\pi}{2}\right)\right)$ is . $\qquad$
-1. $\sqrt{\frac{5 \sqrt{5}-11}{2}}$.
Let $x=\cos \alpha\left(\alpha \in\left(0, \frac{\pi}{2}\right)\right)$. Then $x \in(0,1)$.
Given $y=\frac{x \sqrt{1-x^{2}}}{1+x}$, we have
$$
y'=\frac{1-x-x^{2}}{\sqrt{1-x^{2}}(1+x)} \text {. }
$$

Let $y'=0$. Solving this, we get $x=\frac{\sqrt{5}-1}{2}$ (discard the negative root).
Thus, $y_{\max }=\sqrt{\frac{5 \sqrt{5}-11}{2}}$.