If cos−1x−cos−1y2=α then 4x2−4xycosα+y2 is equal to
−4sin2α
4sin2α
4
2sin2α
We have,
cos−1x−cos−1y2=α⇒ cos−1xy2+1−x21−y24=α⇒ xy2+1−x21−y24=cosα⇒ cosα−xy22=1−x21−y24⇒ cos2α+x2y24−xycosα=1−x2−y24+x2y24⇒ x2+y24−xycosα=sin2α⇒ 4x2−4xycosα+y2=4sin2α