Question

prove this question plzzz

Answer 1

Hiii. .....friend

The answer is here,

Let theta be alpha for my convenience

${sin}^{6} \alpha + {cos}^{6} \alpha$

$= > \: { {(sin}^{2} \alpha )}^{3} + { ({cos}^{2} \alpha )}^{3}$

$= > \: ( {sin}^{2} \alpha + {cos \alpha }^{2} )( {sin}^{4} \alpha + {cos}^{4} \alpha - {cos}^{2} \alpha . {sin}^{2} \alpha )$

$= > \: { ({sin}^{2} \alpha )}^{2} + { {(cos}^{2} \alpha ) }^{2} - {sin}^{2} \alpha . {cos}^{2} \alpha$

$= > \: { ({sin}^{2} \alpha + {cos}^{2} \alpha ) }^{2} - 2 {sin}^{2} \alpha . {cos}^{2} \alpha - {sin}^{2} \alpha . {cos}^{2} \alpha$

$= > \: 1 - 3 {sin}^{2} \alpha . {cos}^{2} \alpha$

LHS =RHS.

:-)hope it helps u.