$\text{ If }{\cos }^{-1}\mathrm{x}-{\cos }^{-1}\left(\frac{\mathrm{y}}{2}\right)=\mathrm{\alpha }\text{ then }4{\mathrm{x}}^2-4\mathrm{xycos\alpha }+{\mathrm{y}}^2=$

$$\begin{gathered} {\cos }^{-1}x=\alpha +{\cos }^{-1}\left(\frac{y}{2}\right) \\ \Rightarrow x=\cos (\alpha +{\cos }^{-1}(\frac{y}{2}\left)\right) \\ \Rightarrow x=\cos \alpha \cos \left({\cos }^{-1}\right(\frac{y}{2})-\sin \alpha \sin ({\cos }^{-1}\left(\frac{y}{2}\right)) \\ \Rightarrow x=\cos \alpha (\frac{y}{2})-\sin \alpha \sqrt{\frac{1-y^2}{4}} \\ \Rightarrow 2x-y\cos \alpha =-\sin \alpha \sqrt{4-y^2} \\ squaring on both sides \\ \Rightarrow 4x^2-4xy\cos \alpha +y^2{\cos }^2\alpha =(4-y^2){\sin }^2\alpha \\ \Rightarrow 4x^2-4xy\cos \alpha +4{\cos }^2\alpha +4{\sin }^2\alpha =4{\sin }^2\alpha \\ \Rightarrow 4x^2-4xy\cos \alpha +4=4{\sin }^2\alpha \end{gathered}$$