已知方程 $x^{2}+4 a x+3 a+1=0(a>1)$ 的两根为 $\tan \alpha 、 \tan \beta$, 且 $\alpha, \beta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, 则 $\tan \frac{\alpha+\beta}{2}$ 的值是( )
$\because\left\{\begin{array}{l}\tan \alpha+\tan \beta=-4 a<0 \\ \tan \alpha \cdot \tan \beta=3 a+1>0\end{array}\right.$,

$\therefore \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{4}{3}$,

$\because \tan \alpha<0, \tan \beta<0, \quad \therefore\left\{^{-\frac{\pi}{2}<\alpha<0}\right.$,

$$
-\frac{\pi}{2}<\beta<0
$$

$\therefore-\pi<\alpha+\beta<0, \quad \therefore-\frac{\pi}{2}<\frac{\alpha+\beta}{2}<0$,

$\because \tan (\alpha+\beta)=\frac{2 \tan \frac{\alpha+\beta}{2}}{1-\tan ^{2} \frac{\alpha+\beta}{2}}=\frac{4}{3}, \therefore \tan \frac{\alpha+\beta}{2}=-2$, 故选 B.