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