If $1 + sin^2 θ – 3sinθ cosθ = 0$, then the value of $cotθ$ is:

2

We are given :-

1 + sin²θ - 3sinθ.cosθ = 0

1 + sin²θ = 3sinθ.cosθ

{ put sin²θ+ cos²θ = 1 }

sin²θ+ cos²θ + sin²θ = 3sinθ.cosθ

cos²θ + 2sin²θ = 3sinθ.cosθ

cos²θ + sin²θ - 2sinθ.cosθ + sin²θ =  sinθ.cosθ

( cosθ -sinθ)² = sinθ( cosθ - sinθ)

cosθ -sinθ = sinθ

2sinθ = cosθ

ATQ,

cotθ = 2