Question

Prove that : ${\cos }^{4}A-{\cos }^{2}A={\sin }^{4}A-{\sin }^{2}A$

Answer 1

${\cos }^{4}A-{\cos }^{2}A={\sin }^{4}A-{\sin }^{2}A$

$LHS.$

$={\cos }^{2}A(1-{\cos }^{2}A)$

$={\cos }^{2}A{\sin }^{2}A[\because {\sin }^{2}A+{\cos }^{2}A=1]$

$RHS$

$={\sin }^{2}A(1-{\sin }^{2}A)$

$={\sin }^{2}A{\cos }^{2}A[\because {\sin }^{2}A+{\cos }^{2}A=1]$

$={\cos }^{2}A{\sin }^{2}A$

$LHS=RHS$