6. Two regular triangular pyramids with the same base are inscribed in the same sphere. It is known that the side length of the base of the two regular triangular pyramids is $a$, and the radius of the sphere is $R$. Let the angles between the lateral faces and the base of the two regular triangular pyramids 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 have $P H \cdot Q H=A H^{2}$,

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 $\tan (\alpha+\beta)=-\frac{4 \sqrt{3} R}{3 a}$.