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

detailed solution

Correct option is C

From the given results 

$\begin{array}{l}\cos x{\sin }^2\beta \cos \alpha -{\cos }^2\alpha {\sin }^2\beta \\ ={\sin }^2\alpha \cos \beta \cos x-{\sin }^2\alpha {\cos }^2\beta \\ \Rightarrow \cos x\left[\cos \alpha {\sin }^2\beta -{\sin }^2\alpha \cos \beta \right]\\ ={\cos }^2\alpha {\sin }^2\beta -{\sin }^2\alpha {\cos }^2\beta \end{array}$

$\begin{array}{l}\Rightarrow \cos x=\frac{{\cos }^2\alpha \left[1-{\cos }^2\beta \right]-\left(1-{\cos }^2\alpha \right){\cos }^2\beta }{\cos \alpha \left[1-{\cos }^2\beta \right]-\left[1-{\cos }^2\alpha \right]\cos \beta }\\ =\frac{{\cos }^2\alpha -{\cos }^2\beta }{(\cos \alpha -\cos \beta )(\cos \alpha \cos \beta +1)}=\frac{\cos \alpha +\cos \beta }{1+\cos \alpha \cos \beta }\\ \Rightarrow \frac{\cos x}{1}=\frac{\cos \alpha +\cos \beta }{1+\cos \alpha \cos \beta }\\ \Rightarrow \frac{1-\cos x}{1+\cos x}=\frac{1+\cos \alpha \cos \beta -\cos \alpha -\cos \beta }{1+\cos \alpha \cos \beta +\cos \alpha +\cos \beta }\\ =\frac{(1-\cos \alpha )(1-\cos \beta )}{(1+\cos \alpha )(1+\cos \beta )}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1-\cos x}{1+\cos x}=\frac{1+\cos \alpha \cos \beta -\cos \alpha -\cos \beta }{1+\cos \alpha \cos \beta +\cos \alpha +\cos \beta }\\ =\frac{(1-\cos \alpha )(1-\cos \beta )}{(1+\cos \alpha )(1+\cos \beta )}\\ \Rightarrow {\tan }^2\frac{x}{2}={\tan }^2\frac{\alpha }{2}\cdot {\tan }^2\frac{\beta }{2}\Rightarrow \tan \frac{x}{2}=\pm \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\\ \Rightarrow \tan \frac{x}{2}=\pm \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\end{array}$