1. If $\theta$ is a real number, and $\frac{1+\sin \theta}{1+\cos \theta}<\frac{\cos \theta}{1-\sin \theta}$, then the range of $y=\sin \frac{\theta}{2}-\cos \frac{\theta}{2}$ is $\qquad$ (express the range as an inequality involving $y$).
$\begin{array}{l}\text { II. } 1 .(-\sqrt{2}, 0) \cup(0, \sqrt{2}) . \\ \frac{1+\sin \theta}{1+\cos \theta}>0\end{array}\right.\right. \\ \Leftrightarrow 2 k \pi-\frac{\pi}{2}<\theta<2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z}) \\ \Leftrightarrow k \pi-\frac{\pi}{4}<\frac{\theta}{2}<k \pi+\frac{\pi}{4} . \\ \text { and } y=\sqrt{2} \sin \left(\frac{\theta}{2}-\frac{\pi}{4}\right), \\ k \pi-\frac{\pi}{2}<\frac{\theta}{2}-\frac{\pi}{4}<k \pi, \\ \text { hence } y \in(-\sqrt{2}, 0) \cup(0, \sqrt{2}) .\end{array}$