\[ \frac{\sinθ}{x} = \frac{\cosθ}{y} \Rightarrow \frac{\sinθ}{\cosθ} = \frac{x}{y} \Rightarrow \tanθ = \frac{x}{y} \Rightarrow \tan^2θ = \frac{x^2}{y^2} \Rightarrow 1 + \tan^2θ = 1 + \frac{x^2}{y^2} \Rightarrow \sec^2θ = \frac{x^2 + y^2}{y^2} \Rightarrow \secθ = \frac{\sqrt{x^2 + y^2}}{y} \Rightarrow \cosθ = \frac{y}{\sqrt{x^2 + y^2}} \] Now, substituting \(\cosθ = \frac{y}{\sqrt{x^2 + y^2}}\) into the original equation: \[ \frac{\sinθ}{x} = \frac{\frac{y}{\sqrt{x^2 + y^2}}}{y} \Rightarrow \frac{\sinθ}{x} = \frac{1}{\sqrt{x^2 + y^2}} \Rightarrow \sinθ = \frac{x}{\sqrt{x^2 + y^2}} \] Therefore: \[ \sinθ - \cosθ = \frac{x}{\sqrt{x^2 + y^2}} - \frac{y}{\sqrt{x^2 + y^2}} = \frac{x - y}{\sqrt{x^2 + y^2}} \quad \text{(Proved)} \]