If $\alpha$and $\beta$ are complementary angles such that $q\sin \alpha =p$ , then the value of $\sin \alpha \cos \beta -\cos \alpha \sin \beta$ is

$\sin \alpha =\frac{p}{q}\Rightarrow \alpha +\beta =90°$

$\Rightarrow \sin \alpha =\sin (90°-\beta )=\cos \beta$

$\Rightarrow \cos \alpha =\cos (90°-\beta )=\sin \beta$

$\cos \alpha =\frac{\sqrt{q^2-p^2}}{q}$

$\therefore$$\sin \alpha \cos \beta -\cos \alpha \sin \beta$

$\Rightarrow {\sin }^2\alpha -{\cos }^2\alpha =\frac{p^2}{q^2}-\frac{(q^2-p^2)}{q^2}=\frac{(2p^2-q^2)}{q^2}$