Two acute angles satisfy $A+B=90^{\circ }$.
Simplify ${\sin }^{2}A+{\sin }^{2}B$.

Since $B=90^{\circ }-A$, we have $\sin B=\sin (90^{\circ }-A)=\cos A$. So ${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}A+{\cos }^{2}A=1$. Answering $2{\sin }^{2}A$ treats $\sin B$ as equal to $\sin A$, which would need $A=B=45$ deg — and even then the value is 1, not 2. Answering 2 additionally forgets that each squared sine is at most 1.