4. Given a tetrahedron $P-ABC$ with three lateral edges $PA, PB, PC$ that are pairwise perpendicular, let the dihedral angles $P-AB-C, P-BC-A$, $P-CA-B$ be $\alpha, \beta, \gamma$ respectively,
then $\frac{\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma}{\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma}=$ $\qquad$ .
The projection of point $P$ on plane $ABC$ is the orthocenter of $\triangle ABC$, then $\sin \alpha=\frac{PC}{CD} \Rightarrow \sin ^{2} \alpha=\frac{PC^{2}}{CD^{2}}$ $=\frac{CH}{CD}=\frac{CH \cdot AB}{CD \cdot AB}=\frac{S_{\triangle HAC}+S_{\triangle HBC}}{S_{\triangle ABC}}$,
thus $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma$
$$
=\frac{2\left(S_{\triangle HAC}+S_{\triangle HBC}+S_{\triangle HAB}\right)}{S_{\triangle ABC}}=2,
$$

therefore $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma$
$$
=3-\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)=1 \text {. }
$$

Hence, $\frac{\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma}{\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma}=2$.