Question

Sin^6A+cos^6A=1-3sin^2A.cos^2A

Answer 1

Consider LHS: sin6A+cos6A = (sin2A)3 + (cos2A)3 = (sin2A + cos2A)3 − 3 sin2A cos2A(sin2A + cos2A) [Since, (a+b)3 = a3 + b3 +3ab(a+b)] = (1)3 − 3 sin2A cos2A(1) [Since, sin2A + cos2A = 1] = 1 − 3 sin2A cos2A = RHS

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