If $\frac{\cos x-\cos \alpha }{\cos x-\cos \beta }=\frac{{\sin }^2\alpha \cos \beta }{{\sin }^2\beta \cos \alpha }$ then cos x=

The given relation can be written as $\cos x({\sin }^2\beta \cos \alpha -{\sin }^2\alpha \cos \beta )$$={\cos }^2\alpha {\sin }^2\beta -{\sin }^2\alpha {\cos }^2\beta$

$\Rightarrow \cos x=\frac{{\cos }^2\alpha (1-{\cos }^2\beta )-(1-{\cos }^2\alpha ){\cos }^2\beta }{\cos \alpha (1-{\cos }^2\beta )-\cos \beta (1-{\cos }^2\alpha )}$

$=\frac{{\cos }^2\alpha -{\cos }^2\beta }{(\cos \alpha -\cos \beta )(1+\cos \alpha \cos \beta )}=\frac{\cos \alpha +\cos \beta }{1+\cos \alpha \cos \beta }$