Question

cos 4a-cos2a=sin4a-sin2a

Answer 1

cos 4 A - Cos2A

= (cos2A)2 - (cosA)2

= (1- sin2A)2 - cos2A

= 1 + sin4A - 2sin2A  -( 1- sin2A)

=1+ sin4A - 2 sin2A -1 + sin2A

=sin4A - sin2A

since LHS = RHS 

hence proved.

Answer 2

To prove:

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

Solution:

Consider the LHS,

$\cos ^{4} A-\cos ^{2} A=\left(\cos ^{2} A\right)^{2}-(\cos A)^{2}$

By formula, $\bold{\sin ^{2} \theta+\cos ^{2} \theta=1}$

$\cos ^{2} \theta=1-\sin ^{2} \theta$ Substitute in the above equation, we get

$\cos ^{4} A-\cos ^{2} A=\left(1-\sin ^{2} A\right)^{2}-(\cos A)^{2}$

$=1+\sin ^{4} A-2 \sin ^{2} A-\left(1-\sin ^{2} A\right)$

$=1+\sin ^{4} A-2 \sin ^{2} A-1+\sin ^{2} A$

$=\sin ^{4} A-\sin ^{2} A=\mathrm{RHS}$

Hence proved.