If 2cot θ = 3, then $\frac{\sqrt{13}cosθ-3tanθ}{3tanθ+\sqrt{13}sinθ}$ is :

$\frac{1}{4}$

Given :-

2cot θ = 3

cot θ = \(\frac{3 }{2}\)

{ we know cot θ = \(\frac{B}{P}\) }

According to pythagoras theorem,

P² + B² = H²

2² + 3² = H²

H = √13

Now,

$\frac{\sqrt{13}cosθ-3tanθ}{3tanθ+\sqrt{13}sinθ}$

= $\frac{\sqrt{13} × 3/√13 - 3× 2/3}{ 3× 2/3+\sqrt{13}× 2/√13 }$

= $\frac{1}{4 }$