$If\text{\hspace{0.17em}\hspace{0.17em}}\cos y=x\cos \left(a+y\right)\text{\hspace{0.17em}\hspace{0.17em}}and\frac{dy}{dx}=\frac{k}{1+x^2-2x\cos a},thenthevalueofkis$

$$\begin{gathered} \cos y=x\cos y\cos a-x\sin y \sin a\Rightarrow x\sin y\sin a=\cos y(x\cos a-1) \\ \Rightarrow \tan y=\frac{x\cos a-1}{x\sin a} \\ \Rightarrow se{c}^{2}y \frac{dy}{dx}=\frac{\sin a}{x^2{\sin }^{2}a} \end{gathered}$$
$\Rightarrow \frac{dy}{dx}=\frac{{\cos }^{2}y}{x^2\sin a}=\frac{1}{x^2\sin a}\frac{x^2{\sin }^{2}a}{x^2{\sin }^{2}a+{(x\cos a-1)}^2}=\frac{\sin a}{1+x^2-2x\cos a}$

$\Rightarrow k=\sin a$