**Q)** If 1 + sin^{2} θ = 3 sin θ cos θ, then prove that tan θ = 1 or 1/2

**Ans:**

Given that 1 + sin^{2} θ = 3 sin θ cos θ

We know that sin^{2} θ + cos^{2} θ = 1

Hence, 1 + sin^{2} θ = 3 sin θ cos θ will become:

(sin^{2} θ + cos^{2} θ) + sin^{2} θ = 3 sin θ cos θ

2 sin^{2} θ + cos^{2} θ = 3 sin θ cos θ

Let’s divide both sides by cos^{2} θ, we get:

2 tan^{2} θ + 1 = 3 tan θ

2 tan^{2} θ – 3 tan θ + 1 = 0

Let’s consider tan θ = X,

then the above equation will become 2 X^{2} – 3 X + 1 = 0

By solving this equation for its roots, we get:

2 X^{2} – 2X – X + 1 = 0

2 X (X – 1) – (X – 1) = 0

(2 X – 1) (X – 1) = 0

Therefore X = 1 or

Since we had taken X = tan θ,

**Therefore, we get tan θ = 1 and tan θ = **