(iii) Sin6 A + Cos 6A = 1 - 3sin2 A cos2 A.
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sin^2A + cos^2A =1
Cubing on both sides
(sin^2A+cos^2A)^3=1^3
(sin^2A)^³+(cos^2A)^3 + 3sin^2Acos^2A(sin^2A+cos^2A)=1
sin^6A+cos^6A+3sin^2Acos^2A(1)=1
sin^6A+cos^6A=1-3sin^2Acos^2A
Hence, proved.
Aa006raw:
wrong answer bro
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