Q. 50

# If 1+sin2θ = 3 sinθ . cosθ, then prove that tan θ = 1 or 1/2.

Given: 1+sin2θ =3 sin θ cos θ

Divide by cos2 θ to both the sides, we get sec2 θ + tan2 θ = 3 tan θ

1+ tan2 θ+ tan2 θ = 3tan θ

2 tan2 θ –3tan θ +1 = 0

Let tanθ = x

2x2 – 3x +1 = 0

2x2 – 2x – x +1 = 0

2x ( x 1) 1(x 1) = 0

(2x 1)(x 1) = 0

Putting each of the factor = 0, we get

x = 1 or And above, we let tan θ = x Hence Proved

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