4. Let $O$ be the center of the base $\triangle ABC$ of the regular tetrahedron $P-ABC$. The dihedral angles between each pair of lateral edges are $\alpha$, and the angle between $PC$ and the plane $PAB$ is $\beta$. Denote the distance from point $O$ to each face as $d$. A moving plane through point $O$ intersects $PC$ at point $S$, and the extensions of $PA$ and $PB$ at points $Q$ and $R$, respectively. Then $\frac{1}{PQ}+\frac{1}{PR}+\frac{1}{PS}$ $\begin{array}{lllll}= & & & \end{array}$ $\qquad$ (the result should be expressed in terms of $\alpha, \beta, d$).
4. $\frac{\sin \beta}{d}$.

Notice,
$$
\begin{array}{l}
V_{\text {tetrahedron } S-P Q R}=\frac{1}{3} S_{\triangle P Q R} h \\
=\frac{1}{3}\left(\frac{1}{2} P Q \cdot P R \sin \alpha\right) P S \sin \beta .
\end{array}
$$

Then $V_{\text {tetrahedron } S-P Q R}$
$$
\begin{aligned}
= & V_{\text {tetrahedron } O-P Q R}+V_{\text {tetrahedron } O-P R S}+V_{\text {tetrahedron } O-P Q S} \\
= & \frac{1}{3} S_{\triangle P Q R} d+\frac{1}{3} S_{\triangle P R S} d+\frac{1}{3} S_{\triangle P Q S} d \\
= & \frac{d}{3} \cdot \frac{1}{2} P Q \cdot P R \sin \alpha+\frac{d}{3} \cdot \frac{1}{2} P S \cdot P R \sin \alpha+ \\
& \frac{d}{3} \cdot \frac{1}{2} P Q \cdot P S \sin \alpha \\
\Rightarrow & P Q \cdot P R \cdot P S \sin \beta \\
& =d(P Q \cdot P R+P R \cdot P S+P Q \cdot P S) \\
\Rightarrow & \frac{1}{P Q}+\frac{1}{P R}+\frac{1}{P S}=\frac{\sin \beta}{d} .
\end{aligned}
$$