Which expression is equal to ${\tan }^{2}A-{\sin }^{2}A$?

Convert to sin and cos: $\frac{{\sin }^{2}A}{{\cos }^{2}A}-{\sin }^{2}A={\sin }^{2}A\left(\frac{1-{\cos }^{2}A}{{\cos }^{2}A}\right)={\sin }^{2}A\cdot \frac{{\sin }^{2}A}{{\cos }^{2}A}={\tan }^{2}A{\sin }^{2}A$. A difference does not turn into a sum, so ${\tan }^{2}A+{\sin }^{2}A$ is wrong; reversing the order of subtraction only flips the sign. Answering ${\sin }^{4}A$ drops the ${\cos }^{2}A$ denominator, since the tidy form is really $\frac{{\sin }^{4}A}{{\cos }^{2}A}$. Check at $A=60$ deg: $3-0.75=2.25$ and ${\tan }^{2}60^{\circ }{\sin }^{2}60^{\circ }=3\times 0.75=2.25$.