If $x\sin y=\sin (y+a)$ and $\frac{dy}{dx}=\frac{A}{1+x^2-2x\cos a}$ then the value of A is

$x\sin y=\sin y\cos a+\cos y\sin a$

$\Rightarrow \sin y(x-\cos a)=\cos y\sin a$

$\Rightarrow \frac{\sin y}{\cos y}=\frac{\sin a}{x-\cos a}\Rightarrow y={\mathrm{Tan}}^{-1}\left(\frac{\sin a}{x-\cos a}\right)$

$\frac{dy}{dx}=\frac{1}{1+{\left({\displaystyle \frac{\sin \alpha }{x-\cos \alpha }}\right)}^2}\times \frac{-\sin a}{{\left(x-\cos \alpha \right)}^2}=\frac{-\sin a}{x^2-2x\cos a+1}$