Mark the correct alternative in the following:
sin6 A + cos6 A + 3 sin2 A cos2 A =
sin6A + cos6A = (sin2A)3 + (cos2A)3
=(sin2A+ cos2A)(sin4A+cos4A-sin2Acos2A)
= 1 (sin4A + cos4A – sin2Acos2A)
∴ sin4A + cos4A – sin2Acos2A+ 3 sin2 A cos2 A
⇒ sin4A + cos4A + 2sin2Acos2A
⇒ (sin2A + cos2A)2
= 12
=1