$If A-B={60}^0 then S{\mathrm{in}}^{2}A+Si{n}^{2}B-SinASinB =$

$\begin{array}{rcl}\mathrm{A}-\mathrm{B}& =& {60}^0\\ & & {\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}\\ & =& 1-{\cos }^2\mathrm{A}+{\sin }^2\mathrm{B}-\frac{1}{2}\sin \mathrm{A} \sin \mathrm{B}\\ & =& 1-\left({\cos }^2\mathrm{A}-{\sin }^2\mathrm{B}\right)-\frac{1}{2}\left(\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})\right)\\ & =& 1-\cos {60}^0 \cos (\mathrm{A}+\mathrm{B})-\frac{1}{2}\left(\frac{1}{2}\right)+\frac{1}{2}\cos (\mathrm{A}+\mathrm{B})\\ & =& 1-\frac{1}{2}\cos (\mathrm{A}+\mathrm{B})-\frac{1}{4}+\frac{1}{2}\cos (\mathrm{A}+\mathrm{B})\\ & =& 1-\frac{1}{4}=\frac{3}{4}\end{array}$