5. Let $\sin \theta+\cos \theta=\frac{\sqrt{2}}{3}, \frac{\pi}{2}<\theta<\pi$. Then the value of $\operatorname{tg} \theta -\operatorname{ctg} \theta$ is $\qquad$ .
5. $-\frac{8 \sqrt{2}}{7}$.

From $\sin \theta+\cos \theta=\frac{\sqrt{2}}{3}$, we get $\sin \theta \cos \theta=-\frac{7}{18}$.
Also, $\frac{\pi}{2}<\theta<\pi$, so $\sin \theta>0, \cos \theta<0$,
$\sin \theta-\cos \theta$
$=\sqrt{(\sin \theta+\cos \theta)^{2}-4 \sin \theta \cos \theta}=\frac{4}{3}$.
$\therefore \operatorname{tg} \theta-\operatorname{ctg} \theta$
$=\frac{(\sin \theta+\cos \theta)(\sin \theta-\cos \theta)}{\sin \theta \cos \theta}$
$=-\frac{8 \sqrt{2}}{7}$.