A student says $\sin (90^{\circ }-A)=\mathrm{cosec}A$, reasoning that "the co- prefix means you flip it".
What does $\sin (90^{\circ }-A)$ actually equal?

$90^{\circ }-A$ is the COMPLEMENT of A, and in a right triangle the side opposite one acute angle is the side adjacent to the other, so $\sin (90^{\circ }-A)=\cos A$. Flipping a ratio is a different relationship: the reciprocal of $\sin A$ is $\mathrm{cosec}A$. Check at $A=30$ deg: $\sin 60^{\circ }\approx 0.866=\cos 30^{\circ }$, while $\mathrm{cosec}30^{\circ }=2$.