If ${\cos }^{-1}\mathrm{x}-{\cos }^{-1}\frac{\mathrm{y}}{2}=\mathrm{\alpha }$ then $4{\mathrm{x}}^2-4\mathrm{xycos}\mathrm{\alpha }+{\mathrm{y}}^2$ is equal to

$\begin{array}{l}\text{ Given, }{\cos }^{-1}\mathrm{x}-{\cos }^{-1}\frac{\mathrm{y}}{2}=\mathrm{\alpha }\\ \Rightarrow {\cos }^{-1}\left(\frac{\mathrm{xy}}{2}+\sqrt{1-{\mathrm{x}}^2}\sqrt{1-\frac{{\mathrm{y}}^2}{4}}\right)=\mathrm{\alpha }\\ \Rightarrow \frac{\mathrm{xy}}{2}+\sqrt{1-{\mathrm{x}}^2}\sqrt{1-\frac{{\mathrm{y}}^2}{4}}=\cos \mathrm{\alpha }\\ \Rightarrow 2\sqrt{1-{\mathrm{x}}^2}\sqrt{1-\frac{{\mathrm{y}}^2}{4}}=2\cos \mathrm{\alpha }-\mathrm{xy}\end{array}$

On squaring both sides, we get 

$\begin{array}{c}4\left(1-{\mathrm{x}}^2\right)\left(\frac{4-{\mathrm{y}}^2}{4}\right)=4{\cos }^2\mathrm{\alpha }+{\mathrm{x}}^2{\mathrm{y}}^2-4\mathrm{xycos}\mathrm{\alpha }\\ \Rightarrow \left(1-{\mathrm{x}}^2\right)\left(4-{\mathrm{y}}^2\right)=4{\cos }^2\mathrm{\alpha }+{\mathrm{x}}^2{\mathrm{y}}^2-4\mathrm{xycos}\mathrm{\alpha }\\ \Rightarrow 4-{\mathrm{y}}^2-4{\mathrm{x}}^2+{\mathrm{x}}^2{\mathrm{y}}^2=4{\cos }^2\mathrm{\alpha }+{\mathrm{x}}^2{\mathrm{y}}^2-4\mathrm{xycos}\mathrm{\alpha }\\ \Rightarrow 4{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{xycos}\mathrm{\alpha }=-\left(4-4{\cos }^2\mathrm{\alpha }\right)\\ \Rightarrow 4{\mathrm{x}}^2-4\mathrm{xycos}\mathrm{\alpha }+{\mathrm{y}}^2=-4{\sin }^2\mathrm{\alpha }\end{array}$