# If 1+sin2θ =3sinθ cos θ, then prove that tanθ =1 or .

Given: 1+sin2 θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin2 θ,

We get

cot2 θ +1+1 = 3 cot θ [, cosec2 θ – cot2 θ = 1 cosec2 θ = cot2 θ +1]

cot2 θ +2 = 3 cot θ

cot2 θ –3 cot θ +2 = 0

Splitting the middle term and then solving the equation,

cot2 θ – cot θ –2 cot θ +2 = 0

cot θ(cot θ -1)2(cot θ +1) = 0

(cot θ - 1)(cot θ - 2) = 0

cot θ = 1, 2

Hence, proved.

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