Question

[(sin4a*cos2a)-(cos4a*sin2a)]/[cos²a-sin²a]=tan2a

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

Step-by-step explanation:

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.