5. Let $\sin \theta+\cos \theta=\frac{\sqrt{2}}{3}, \frac{\pi}{2}<\theta<\pi$, then the value of $\tan \theta-\cot \theta$ is $\qquad$
$\begin{array}{l}\text { 5. From } \sin \theta+\cos \theta=\frac{\sqrt{2}}{3} \text {, we get } \sin \theta \cdot \cos \theta=-\frac{7}{18} \text {, and } \frac{\pi}{2}<\theta<\pi, \text { so } \sin \theta>0, \cos \theta<0, \text { therefore } \sin \theta-\cos \theta=\sqrt{(\sin \theta+\cos \theta)^{2}-4 \sin \theta \cos \theta}=\frac{4}{3} \text {. } \\ \therefore \tan \theta-\cot \theta=\frac{(\sin \theta+\cos \theta)(\sin \theta-\cos \theta)}{\sin \theta \cdot \cos \theta}=-\frac{8}{7} \sqrt{2} \text {. }\end{array}$