solve 5th no. like this identity,
(sin³A)²+(cos³A)²
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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
Anonymous:
i want to be solved by squared of cubes
Why ??
plz mark it
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Hope it helps !!
Refer to the attachment
#H
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