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
Because $\sqrt{2}+\sqrt{3}>\pi$, we have $0\cos \left(\sqrt{3}-\frac{\pi}{2}\right)$, which means $\sin \sqrt{2}>\sin \sqrt{3}$.

Also, $00, \cos \sqrt{3}0$, the curve represented by the equation is an ellipse.
Since
$$
\begin{aligned}
& (\sin \sqrt{2}-\sin \sqrt{3})-(\cos \sqrt{2}-\cos \sqrt{3}) \\
= & 2 \sqrt{2} \sin \frac{\sqrt{2}-\sqrt{3}}{2} \sin \left(\frac{\sqrt{2}+\sqrt{3}}{2}+\frac{\pi}{4}\right),
\end{aligned}
$$

and $-\frac{\pi}{2}0$, thus (*) is $<0$.
That is, $\sin \sqrt{2}-\sin \sqrt{3}<\cos \sqrt{2}-\cos \sqrt{3}$, so the curve represents an ellipse with foci on the $y$-axis, hence the answer is $\mathrm{C}$.