Example 4. $\cos ^{2} 10^{\circ}+\cos ^{2} 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}=$
Consider the dual form $B=\sin ^{2} 10^{\circ}+$ $\sin ^{2} 50^{\circ}-\cos 40^{\circ} \cos 80^{\circ}$ of the original expression $A=$, then $A+B=2-$ $\cos 40^{\circ}, A-B=\frac{1}{2}+\cos 40^{\circ}$.
Thus, $A=\frac{3}{4}$.