if cot θ = $\dfrac{a}{b}$, then value of $\dfrac{a sin θ - b cos θ}{a sin θ + b cos θ}$ is

The correct answer is 0

$\dfrac{a sin θ - b cos θ}{a sin θ + b cos θ}$
= $\dfrac{a - b cot θ}{a + b cot θ}$
= $\dfrac{a - b × \dfrac{a}{b}}{a + b × \dfrac{a}{b}}$
= $\dfrac{a - a}{a + a}$
= 0