For an acute angle A a student simplifies $\sqrt{{\sec }^{2}A-1}$ and writes $\sec A-1$. What is the correct value?

Since ${\sec }^{2}A-1={\tan }^{2}A$, the root is $\sqrt{{\tan }^{2}A}=\tan A$ for an acute angle. The student square-rooted the two terms separately, which is never allowed. Check at $A=60$ deg: $\sec 60^{\circ }=2$, so $\sqrt{4-1}=\sqrt{3}=\tan 60^{\circ }$, whereas $\sec 60^{\circ }-1=1$.