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

$4sin^2 \alpha $

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

$⇒ cos^{-1}\begin{Bmatrix}\frac{xy}{2}+\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}\end{Bmatrix}= \alpha $

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

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

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

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

$⇒ 4x^2 - 4xy cos \alpha + y^2 = 4 sin^2 \alpha $