7. In $\triangle A B C$, if $\tan \frac{A}{2}+\tan \frac{B}{2}=1$, then the minimum value of $\tan \frac{C}{2}$ is $\qquad$ .
7. $\frac{3}{4}$.

From the given, we know
$$
\begin{array}{l}
\tan \frac{C}{2}=\frac{1}{\tan \frac{A+B}{2}}=\frac{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}} \\
=1-\tan \frac{A}{2} \cdot \tan \frac{B}{2} \\
\geqslant 1-\left(\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{2}\right)^{2} \\
=1-\left(\frac{1}{2}\right)^{2}=\frac{3}{4} .
\end{array}
$$

The equality holds if and only if $\tan \frac{A}{2}=\tan \frac{B}{2}=\frac{1}{2}$.
Therefore, the minimum value of $\tan \frac{C}{2}$ is $\frac{3}{4}$.