Question

8 and 9......................

Answer 1

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8) L.H.S. = Cos6A+Sin6A

= (cos2A)3 + (sin2A)3

= (cos2A + sin2A)3 -3cos2Asin2A(cos2A + sin2A)

= 13 -3cos2Asin2A(1)2 [since, sin2A + cos2A = 1]

= 1 - 3cos^2Asin^2A

= R.H.S.

Hence Proved

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9) Sin⁴A-Cos⁴A

=(sin²A)²-(cos²A)²

=(sin²A+cos²A)(sin²A-cos²A)

=sin²A-cos²A [sin²A+cos²A=1]

sin²A+cos²A-2cos²A=1-2cos²A (proved)

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HERE IS YOUR ANSWER,

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