3. Given $\sin \alpha+\cos \alpha=-\frac{3}{\sqrt{5}}$, and $|\sin \alpha|>|\cos \alpha|$, find the value of $\operatorname{tg} \frac{\alpha}{2}$.
3. From the given $\sin \alpha|\cos \alpha|$, we know $\sin \alpha < \cos \alpha$. From $\sin \alpha + \cos \alpha = -\frac{3}{\sqrt{5}}$, we get $\sin \alpha - \cos \alpha = -\frac{1}{\sqrt{5}}$. Therefore, $\sin \alpha = -\frac{2}{\sqrt{5}}, \cos \alpha = -\frac{1}{\sqrt{5}}$, hence $\operatorname{tg} \frac{\alpha}{2} = \frac{1 - \cos \alpha}{\sin \alpha} = -\frac{\sqrt{5} + 1}{2}$.