6. Two regular tetrahedra with the same base are inscribed in the same sphere. It is known that the side length of the base of the two regular tetrahedra is $a$, and the radius of the sphere is $R$. Let the angles between the side faces and the base of the two regular tetrahedra be $\alpha$ and $\beta$, respectively. Then the value of $\tan (\alpha+\beta)$ is $\qquad$
6. $-\frac{4 \sqrt{3} R}{3 a}$.

As shown in Figure 3, let $P-A B C$ and $Q-A B C$ be two regular tetrahedrons with the same base, both inscribed in a sphere $O$ with radius $R$. Connect $P Q$, then $P Q \perp$ plane $A B C$, and the foot of the perpendicular is $H$, and $P Q$ passes through the center of the sphere $O$. Connect $C H$ and extend it to intersect $A B$ at $D$, then $D$ is the midpoint of $A B$, and $C D \perp A B$. It is easy to see that $\angle P D H=\alpha, \angle Q D H=\beta$.
Since $D H=\frac{1}{3} C D=\frac{\sqrt{3} a}{6}$,
we have $P H=D H \cdot \tan \alpha=\frac{\sqrt{3} a}{6} \cdot \tan \alpha$.
Similarly, $Q H=\frac{\sqrt{3} a}{6} \cdot \tan \beta$.
From $P H+Q H=2 R$, we get
$\tan \alpha+\tan \beta=\frac{4 \sqrt{3} R}{a}$.
Also, since $\angle P A Q=90^{\circ}, A H \perp P Q$, by the projection theorem, we get $P H \cdot Q H=A H^{2}$,

i.e., $\left(\frac{\sqrt{3} a}{6}\right)^{2} \cdot \tan \alpha \cdot \tan \beta=\left(\frac{\sqrt{3} a}{3}\right)^{2}$.
Thus, $\tan \alpha \cdot \tan \beta=4$.
From (1) and (2), we get $\tan (\alpha+\beta)=-\frac{4 \sqrt{3} R}{3 a}$.