5. The curve represented by the equation $\frac{x^{2}}{\sin \sqrt{2}-\sin \sqrt{3}}+\frac{y^{2}}{\cos \sqrt{2}-\cos \sqrt{3}}=1$ is ( ).
(A) an ellipse with foci on the $x$-axis
(B) a hyperbola with foci on the $x$-axis
(C) an ellipse with foci on the $y$-axis
(D) a hyperbola with foci on the $y$-axis
5.C.

From $\sqrt{2}+\sqrt{3}>\pi$, we have $0\cos \left(\sqrt{3}-\frac{\pi}{2}\right)$,
which means $\sin \sqrt{2}>\sin \sqrt{3}$.
Also, $00 \cdot \cos \sqrt{3}$ $0$, so the curve represented by the equation is an ellipse. And
$$
\begin{array}{l}
(\sin \sqrt{2}-\sin \sqrt{3})-(\cos \sqrt{2}-\cos \sqrt{3}) \\
=2 \sqrt{2} \sin \frac{\sqrt{2}-\sqrt{3}}{2} \cdot \sin \left(\frac{\sqrt{2}+\sqrt{3}}{2}+\frac{\pi}{4}\right) .
\end{array}
$$

Since $-\frac{\pi}{2}0$. Therefore, expression (1) $<0$.
Thus, $\sin \sqrt{2}-\sin \sqrt{3}<\cos \sqrt{2}-\cos \sqrt{3}$. Hence, the curve represents an ellipse with foci on the $y$-axis.