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Question
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
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Solution
L.H.S. = sin4θ - cos4θ
L.H.S. = (sin2θ)2 - (cos2θ)2
L.H.S. = (sin2θ - cos2θ)(sin2θ + cos2θ)
L.H.S. = (sin2θ - cos2θ) x 1
L.H.S. = sin2θ - cos2θ
L.H.S. = R.H.S.
L.H.S.= sin2θ - (1 - sin2θ)
L.H.S. = sin2θ - 1 + sin2θ
L.H.S. = 2sin2θ - 1
L.H.S. = R.H.S
L.H.S. = 2(1 - cos2θ) - 1
L.H.S. = 2 - 2cos2θ - 1
L.H.S. = 1 - 2cos2θ
L.H.S. = R.H.S.
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