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Q.
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
Conic Sections
Solution:
Since $\sqrt{2}+\sqrt{3}>\pi$, so $0<\frac{\pi}{2}-\sqrt{2}<\sqrt{3}-\frac{\pi}{2}<\frac{\pi}{2}$
and $\cos \left(\frac{\pi}{2}-\sqrt{2}\right)>\cos \left(\sqrt{3}-\frac{\pi}{2}\right)$, i.e. $\sin \sqrt{2}>\sin \sqrt{3}$.
Since $0<\sqrt{2}<\frac{\pi}{2}, \frac{\pi}{2}<\sqrt{3}<\pi$, so $\cos \sqrt{2}>0, \cos \sqrt{3}<0$, and $\cos \sqrt{2}-\cos \sqrt{3}>0$.
Thus the curve represented by the equation is an ellipse.
Since $(\sin \sqrt{2}-\sin \sqrt{3})-(\cos \sqrt{2}-\cos \sqrt{3})=2 \sqrt{2} \sin \frac{\sqrt{2}-\sqrt{3}}{2} \sin \left(\frac{\sqrt{2}+\sqrt{3}}{2}+\frac{\pi}{4}\right)$
and $ \frac{-\pi}{2}<\frac{\sqrt{2}-\sqrt{3}}{2}< 0$
we get $\sin \frac{\sqrt{2}-\sqrt{3}}{2}<0, \frac{\pi}{2}<\frac{\sqrt{2}+\sqrt{3}}{2}<\frac{3 \pi}{4}, \frac{3 \pi}{4}<\frac{\sqrt{2}+\sqrt{3}}{2}+\frac{\pi}{4}<\pi, \sin \left(\frac{\sqrt{2}+\sqrt{3}}{2}+\frac{\pi}{4}\right)>0$ so the expression $(*)$ is less than 0 .
That is $\sin \sqrt{2}-\sin \sqrt{3}<\cos \sqrt{3}-\cos \sqrt{3}$, therefore the curve is an ellipse with foci on the $y$-axis.