Question

rhs prove ......plzz solve fast

Answer 1

$lhs = { \sin}^{6} \theta + { \cos}^{6} \theta \\ = {( { \sin }^{2} \: \theta)}^{3} + {({ \ \cos }^{2} \theta)}^{3} \\ = ({ \sin }^{2} \theta \: +{\cos }^{2} \theta)( ( {{\sin }^{2} \theta)}^{2} +({{\cos }^{2} \theta)}^{2} - 2{{\sin}^{2} \theta \cos }^{2} \theta) \\ = 1( ( {{\sin }^{2} \theta)}^{2} +({{\cos }^{2} \theta)}^{2} - 2{{\sin}^{2} \theta \cos }^{2} \theta) \\ = ({\sin }^{2} \theta+{\cos }^{2} \theta)^{2} -{{\sin}^{2} \theta \cos }^{2} \theta - 2{{\sin}^{2} \theta \cos }^{2} \theta\\ = 1 - 3{{\sin}^{2} \theta \cos }^{2} \theta \\ = rhs$

Answer 2

hiii mate here is ur solution