9. If the equation $\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma$ holds, what algebraic relationship should $\alpha, \beta, \gamma$ satisfy?
9. $\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma \Rightarrow \tan \alpha+\tan \beta=-\tan \gamma(1-\tan \alpha \tan \beta)$, if $1-$ $\tan \alpha \tan \beta=0$, then $\tan \alpha+\tan \beta=0$. On the other hand, $1-\tan \alpha \tan \beta=0 \Leftrightarrow \tan \alpha \tan \beta=1>0 \Rightarrow \tan \alpha$ and $\tan \beta$ have the same sign $\Rightarrow \tan \alpha+\tan \beta \neq 0$, contradiction. Therefore, $1-\tan \alpha \tan \beta \neq 0$, so we have $\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=-\tan \gamma=$ $\tan (-\gamma)$, i.e., $\tan (\alpha+\beta)=\tan (-\gamma)$. Thus, $(\alpha+\beta)-(-\gamma)=k \pi \Leftrightarrow \alpha+\beta+\gamma=k \pi$. Conversely, if $\alpha+\beta+\gamma=k \pi$, and $\alpha$, $\beta$, $\gamma$ are not odd multiples of $\frac{\pi}{2}$, then $\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma$ holds.