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If A=30o, verify that i) sin2A = 2sinAcosA ii) cos2A = 1−tan2A1+tan2A
1 Answer
written 5.6 years ago by |
Solution:
I) L.H.S = sin2A
= sin2(30o)
= 2sin60o
= √32
R.H.S = 2Sin2CosA
=2Sin300Cos30o
=2(12)(√32)
=√32
hence, SIn2A = 2SinACosA
ii) L.H.S = cos2A=cos2(30o)
=cos60o
=12
R.H.S = 1−tan2A1+tan2A
=1−tan230o1+tan230o
=1−(1√3)21+(1√3)2
=12
hence, Cos2A=1−tan2A1+tan2A