$\text{If }\mathrm{A}+\mathrm{B}+\mathrm{C}={180}^{\circ }\text{ then }{\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}+{\sin }^2\mathrm{C}=$

$\begin{array}{l}={\sin }^2\mathrm{A}+1-{\cos }^2\mathrm{B}+{\sin }^2\mathrm{C}=1-\left({\cos }^2-{\sin }^2\mathrm{A}\right)+{\sin }^2\mathrm{C}\\ =1-\cos (\mathrm{A}+\mathrm{B})\cos (\mathrm{A}-\mathrm{B})+1-{\cos }^2\mathrm{C}=2+\cos \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})+\cos (\mathrm{A}+\mathrm{B}\left)\right]\\ =2(1+\cos \mathrm{Acos}\mathrm{Bcos}\mathrm{C})\end{array}$