sin⁴A - cos⁴A = 2sin²A - 1

1.	sin⁴A - cos⁴A = 2sin²A - 1
Answer» We have,L.H.S = sin4A - cos4A{tex}\\Rightarrow{/tex}\xa0L.H.S = (sin2A)2 - (cos2A)2{tex}\\Rightarrow{/tex}\xa0L.H.S = (sin2A + cos2A) (sin2A - cos2A) {tex}\\left[\\because a^2-b^2=\\left(a+b\\right)\\left(a-b\\right)\\right]{/tex}{tex}\\Rightarrow{/tex}\xa0L.H.S = sin2A - cos2 A [\xa0{tex}\\therefore{/tex}\xa0sin2 A + cos2 A = 1]{tex}\\Rightarrow{/tex}\xa0L.H.S = (1 - cos2A)-\xa0cos2 A = 1 - 2cos2 A = 1-2(1-sin2A) = 2sin2A-1 = R.H.S

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