Some tricky trigonometric problem

Question

If $\frac{\sinθ}{x}=\frac{\cosθ}{y}$ , then $\sin(θ)-\cos(θ)=?$

Answer 1

For $x\ne y,$ $$\dfrac{\sin\theta}x=\dfrac{\cos\theta}y=\dfrac{\sin\theta-\cos\theta}{x-y}$$

Again, $$\dfrac{\sin\theta}x=\dfrac{\cos\theta}y=\pm\sqrt{\dfrac{\sin^2\theta+\cos^2\theta}{x^2+y^2}}=?$$

Answer 2

$\tan \theta =\frac x y$ so $\sin \theta -\cos \theta =cos \theta (tan \theta -1)=\pm \frac {\frac x y -1} {\sqrt {1+\frac {x^{2}} {y^{2}}}}$ (since $\cos \theta =\frac 1 {\sec \theta}$ and $\sec \theta =\pm \sqrt {1+\tan^{2} \theta}$.