Let $\alpha$ and $\beta$ be the roots of

$x^{2}+(2-\tan \theta) x-$ $(1+\tan \theta)=0$

$\alpha+\beta=\tan \theta-2 $           ………$(1)$

$\alpha \beta=-\tan \theta-1$           ……..$(2)$

$(1)+(2) \Rightarrow \alpha+\beta+\alpha \beta=-3 \Rightarrow(\alpha+1)(\beta+1)=-2$

Hence either $\alpha+1=-2 $ and $ \beta+1=1$ or

$\alpha+1=-1 $ and $ \beta+1=2$

i.e. either $\alpha=-3 $ and $ \beta=0$ or $\alpha=-2 $ and $ \beta=1$

i.e. either $\tan \theta=-1$ or $\tan \theta=1$

i.e. either $\theta=\frac{3 \pi}{4}, \frac{7 \pi}{4}$ or $\theta=\frac{\pi}{4}, \frac{5 \pi}{4}$

Hence sum of all possible values of $\theta$ is

$\frac{\pi}{4}+\frac{3 \pi}{4}+\frac{5 \pi}{4}+\frac{7 \pi}{4}=\frac{16 \pi}{4}=4 \pi$