If  ${\int }_0^{\alpha }\frac{dx}{1-\cos \alpha \cos x}=\frac{A}{\sin \alpha }+B(\alpha \ne 0)$ . Then possible of A and B are 

 $$\begin{gathered} I={\int }_0^{\alpha }\frac{dx}{1-\cos \alpha \cos x} \\ I={\int }_0^{\alpha }\frac{dx}{({\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2})-\cos \alpha ({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})} \\ ={\int }_0^{\alpha }\frac{dx}{(1-\cos \alpha ){\cos }^2\frac{x}{2}+(1+\cos \alpha ){\sin }^2\frac{x}{2}} \\ ={\int }_0^{\alpha }\frac{dx}{2{\sin }^2\frac{\alpha }{2}{\cos }^2\frac{x}{2}+2{\cos }^2\frac{\alpha }{2}{\sin }^2\frac{x}{2}} \\ =\frac{1}{2}{\int }_0^{\alpha }\frac{\left({\sec }^2\frac{\alpha }{2}\right){\sec }^2\frac{x}{2}}{{\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{x}{2}}dx \end{gathered}$$
Put  $\tan \frac{x}{2}=t,$  we have  
 $\text{I}={\int }_0^{\tan \frac{\alpha }{2}}\frac{{\sec }^2\frac{\alpha }{2}dt}{{\mathrm{t}}^2+{\tan }^2\frac{\alpha }{2}}dx$
$={\sec }^2\frac{\alpha }{2}\cot \frac{\alpha }{2}{\left[{\tan }^{-1}\left(\frac{t}{\tan }\frac{\alpha }{2}\right)\right]}_0^{\tan \frac{\alpha }{2}}$ 
 $=\frac{2}{\sin \alpha }.\frac{\pi }{4}=\frac{\pi }{2\sin \alpha }$
Thus  $\frac{A}{\sin \alpha }+B=\frac{\pi }{2\sin \alpha }.A=\frac{\pi }{2},B=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}A=}\frac{\pi }{4},$
$B=\frac{\pi }{4\sin \alpha }$  satisfy the last equation.