7. In $\triangle A B C$, if $a+c=2 b$, then $\tan \frac{A}{2} \cdot \tan \frac{C}{2}=$
7. $\frac{1}{3}$.

By the Law of Sines, $a=2 k \sin A, b=2 k \sin B, c=2 k \sin C$, then $\sin A+\sin C=2 \sin B$. Since $B=\pi-(A+C)$, we have $\sin A+\sin C=2 \sin (A+C), 2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}=4 \sin \frac{A+C}{2} \cos \frac{A+C}{2}$. Thus, $\cos \frac{A-C}{2}=2 \cos \frac{A+C}{2}, \cos \frac{A}{2} \cos \frac{C}{2}+\sin \frac{A}{2} \sin \frac{C}{2}=2 \cos \frac{A}{2} \cos \frac{C}{2}-2 \sin \frac{A}{2} \sin \frac{C}{2}$, which simplifies to $\cos \frac{A}{2} \cos \frac{C}{2}=3 \sin \frac{A}{2} \sin \frac{C}{2}$. Therefore, $\tan \frac{A}{2} \cdot \tan \frac{C}{2}=\frac{1}{3}$.