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

Question

$2\sin 2\alpha $

$4$

$4{\sin ^2}\alpha $

$-4{\sin ^2}\alpha $

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