If tan θ = $\dfrac{p}{q}$, find the value of $\dfrac{(\text{p sin θ - q cos θ})}{(\text{p sin θ + q cos θ)}}$.

Given,

$\dfrac{(\text{p sin θ - q cos θ})}{(\text{p sin θ + q cos θ)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \dfrac{\dfrac{(\text{p sin θ - q cos θ})}{\text{cos θ}}}{\dfrac{(\text{p sin θ + q cos θ)}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{p sin θ}}{\text{cos θ}} - \dfrac{\text{q cos θ}}{\text{cos θ}}}{\dfrac{\text{p sin θ}}{\text{cos θ}} + \dfrac{\text{q cos θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\text{p tan θ - q}}{{\text{p tan θ + q}}} \space [\because \text{tan θ} = \dfrac{\text{sin θ}}{\text{cos θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \dfrac{p \times \dfrac{p}{q} - q}{p \times \dfrac{p}{q} + q} \\[1em] \Rightarrow \dfrac{\dfrac{p^2}{q} - q}{\dfrac{p^2}{q} + q} \\[1em] \Rightarrow \dfrac{\dfrac{p^2 - q^2}{q}}{\dfrac{p^2 + q^2}{q}} \\[1em] \Rightarrow \dfrac{p^2 - q^2}{p^2 + q^2}.$

Hence, $\dfrac{(\text{p sin θ - q cos θ})}{(\text{p sin θ + q cos θ)}} = \dfrac{p^2 - q^2}{p^2 + q^2}.$