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Question
Prove the following:
sin4θ + cos4θ = 1 – 2 sin2θ cos2θ
Solution
LH.S. = sin4θ + cos4θ
= (sin2θ)2 + (cos2θ)2
= (sin2θ + cos2θ)2 – 2sin2θ cos2θ ...[∵ a2 + b2 = (a + b)2 – 2ab]
= (1)2 – 2sin2θ cos2θ
= 1 – 2sin2θ cos2θ
= R.H.S.
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