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Question

If 1+ sin2 A = 3sinAcosA , then prove that tanA=1 or 1/2

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Solution

1+Sin²A= 3SinA Cos A.
Cos²A+Sin²A+Sin²A = 3SinA CosA [ 1 = Sin²+Cos A]
Cos²A+2Sin²A = 3SinA CosA....(1)
DIVIDE (1) BY COS²A we get,
1+2Tan²A = 3 TanA
1+2Tan²A - 3 TanA = 0
(2TanA-1) ( TanA-1) = 0
2TanA -1 = 0. TanA -1= 0
2TanA = 1. TanA =1
TanA = 1/2. TanA = 1

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