If ${\cos }^{-1}x-{\cos }^{–1}\frac{y}{2}=\alpha ,$ where $-1\le x\le 1, -2\le y\le 2, x\le \frac{y}{2},$ then for all $x, y, 4x^2-4xy\cos \alpha +y^2$ is equal to :

Given ${\cos }^{-1}x-{\cos }^{-1}\frac{y}{2}=\alpha$

${\cos }^{-1}A-{\cos }^{-1}B={\cos }^{-1}\left(AB+\sqrt{1-A^2}\sqrt{1-B^2}\right),$

$\Rightarrow {\cos }^{-1}\left(\frac{xy}{2}+\sqrt{1-x^2}\times \sqrt{1-\frac{y^2}{4}}\right)=\alpha$

$\Rightarrow \frac{xy}{2}+\sqrt{1-x^2}\times \sqrt{1-\frac{y^2}{4}}=\cos \alpha$

$\Rightarrow \cos \alpha -\frac{xy}{2}=\sqrt{1-x^2}\times \sqrt{1-\frac{y^2}{4}}$

Squaring both sides, we get
${\cos }^2\alpha +\frac{x^2{y}^2}{4}-xy\cos \alpha =\left(1-x^2\right)\left(1-\frac{y^2}{4}\right)$

$\Rightarrow {\cos }^2\alpha +\frac{x^2{y}^2}{4}-xy\cos \alpha =1-x^2-\frac{y^2}{4}+\frac{x^2{y}^2}{4}$

$\Rightarrow x^2-xy\cos \alpha +\frac{y^2}{4}=1-{\cos }^2\alpha$

$\Rightarrow x^2-xy\cos \alpha +\frac{y^2}{4}={\sin }^2\alpha$

$\Rightarrow 4x^2-4xy\cos \alpha +y^2=4{\sin }^2\alpha .$