Question

${\sin }^2{62}^{\circ }-{\cos }^2{28}^{\circ }$

Answer 1

${\sin }^2{62}^{\circ }-{\cos }^2{28}^{\circ }$
= ${\sin }^2(90-28{)}^{\circ }-{\cos }^2{28}^{\circ }$
= ${\cos }^2{28}^{\circ }-{\cos }^2{28}^{\circ }$
= $0$