If 3 cot θ = 4, find the value of $\dfrac{\text{5 sin θ - 3 cos θ}}{\text{5 sin θ + 3 cos θ}}$.

Given,

⇒ 3 cot θ = 4

⇒ cot θ = $\dfrac{4}{3}$

$\dfrac{\text{5 sin θ - 3 cos θ}}{\text{5 sin θ + 3 cos θ}} \\[1em] [\text{Dividing numerator and denominator by sin θ}] \\[1em] \Rightarrow \dfrac{\dfrac{\text{(5 sin θ - 3 cos θ)}}{\text{sin θ}}}{\dfrac{\text{(5 sin θ + 3 cos θ)}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{5 sin θ}}{\text{sin θ}} - \dfrac{\text{3 cos θ}}{\text{sin θ}}}{\dfrac{\text{5 sin θ}}{\text{sin θ}} + \dfrac{\text{3 cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\text{5 - 3 cot θ}}{{5 + \text{3 cot θ}}} \space [\because \text{cot θ} = \dfrac{\text{cos θ}}{\text{sin θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \dfrac{5 - 3 \times \dfrac{4}{3}}{5 + 3\times \dfrac{4}{3}} \\[1em] \Rightarrow \dfrac{5 - 4}{5 + 4} \\[1em] \Rightarrow \dfrac{1}{9}.$

Hence, $\dfrac{\text{5 sin θ - 3 cos θ}}{\text{5 sin θ + 3 cos θ}} = \dfrac{1}{9}$