A textbook states the identity $\sqrt{{\sec }^{2}A-1}=\tan A$ but adds the condition 0° <= $A<90$ deg.
Why is that condition needed?

The identity ${\sec }^{2}A=1+{\tan }^{2}A$ holds wherever both ratios are defined, so $\sqrt{{\sec }^{2}A-1}=\sqrt{{\tan }^{2}A}$, which is the NON-NEGATIVE root. Dropping the modulus and writing $\tan A$ is therefore only safe where $\tan A\ge 0$, that is for 0° <= $A<90$ deg. Note that ${\sec }^{2}A-1={\tan }^{2}A$ is a square and so is never negative, and $\sec A$ is defined for every angle except those where $\cos A=0$ — neither of those is the reason.