4. Given $\sin \alpha+\cos \alpha=-\frac{1}{5}(0<\alpha<\pi)$, then $\tan \alpha=$ $\qquad$
4. Fill in $\frac{3}{4}$. Reason: If this problem is squared to get the value of $\sin 2 \alpha$, and then the universal formula is used to find $\tan \alpha$, it involves a bit of calculation, and the value of $\tan \alpha$ obtained has 2 solutions, one of which needs to be discarded. It can be estimated as follows:

From $00$, and $\sin \alpha+\cos \alpha=-\frac{1}{5}$, so $\cos \alpha<0$. Considering $\sin ^{2} \alpha+\cos ^{2} \alpha=1$ and the result is $-\frac{1}{5}$, we can calculate $\sin \alpha=\frac{3}{5}, \cos \alpha=-\frac{4}{5}$, hence $\tan \alpha=-\frac{3}{4}$.