If $n = \dfrac {\cos \alpha}{\cos \beta}, m = \dfrac {\sin \alpha}{\sin \beta}$, then $(m^{2} - n^{2})\sin^{2}\beta$ is

Given,

$(m^2-n^2)\sin ^2\beta$

$=\left ( \dfrac{\sin ^2\alpha }{\sin ^2\beta } -\dfrac{\cos ^2\alpha }{\cos ^2\beta }\right )\sin ^2\beta$

$=\left ( \dfrac{\cos ^2\beta \sin ^2\alpha-\sin ^2\beta\cos ^2\alpha }{\sin ^2\beta \cos ^2\beta} \right )\sin ^2\beta$

$=\sin ^2\alpha-\dfrac{\sin ^2\beta}{\cos ^2\beta }\cos ^2\alpha$

$=1-\dfrac{\cos ^2\alpha }{\cos ^2\beta }$

$=1-n^2$