Q11 of 118 Page 5

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

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