Question

If 1 + sin²θ = 3sinθ cosθ, then prove that tanθ = 1 or ${\displaystyle \frac{1}{2}}$.

Answer 1

Given: 1 + sin² θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin²θ,

We get, 

${\displaystyle \frac{1+{\sin }^2\theta }{{\sin }^2\theta }=\frac{3\sin \theta \cos \theta }{{\sin }^2\theta }}$

${\displaystyle \Rightarrow \frac{1}{{\sin }^2\theta }+1=\frac{3\cos \theta }{\sin \theta }}$

cosec² θ + 1 = 3 cot θ

Since, cosec² θ – cot² θ = 1 

${\displaystyle \Rightarrow }$ cosec² θ = cot² θ + 1

${\displaystyle \Rightarrow }$ cot² θ + 1 + 1 = 3 cot θ

${\displaystyle \Rightarrow }$ cot² θ + 2 = 3 cot θ

${\displaystyle \Rightarrow }$ cot² θ – 3 cot θ + 2 = 0

Splitting the middle term and then solving the equation,

${\displaystyle \Rightarrow }$ cot² θ – cot θ – 2 cot θ + 2 = 0

${\displaystyle \Rightarrow }$ cot θ(cot θ – 1) – 2(cot θ + 1) = 0

${\displaystyle \Rightarrow }$ (cot θ – 1)(cot θ – 2) = 0

${\displaystyle \Rightarrow }$ cot θ = 1, 2

Since,

tan θ = ${\displaystyle \frac{1}{\cot \theta }}$

tan θ = ${\displaystyle 1,\frac{1}{2}}$

Hence proved.