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

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$\begin{aligned} & \cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha \\ \Rightarrow & \cos ^{-1}\left\{\frac{x y}{2}+\sqrt{1-x^{2}} \sqrt{1-\frac{y^{2}}{4}}\right\}=\alpha \\ \Rightarrow & \frac{x y}{2}+\sqrt{\left(1-x^{2}\right)\left(1-\frac{y^{2}}{4}\right)}=\cos \alpha \\ \Rightarrow & \sqrt{\left(1-x^{2}\right)\left(1-\frac{y^{2}}{4}\right)}=\cos \alpha-\frac{x y}{2} \\ \Rightarrow & \left(1-x^{2}\right)\left(-1-\frac{y^{2}}{4}\right)=\cos ^{2} \alpha+\frac{x^{2} y^{2}}{4}-x y \cos \alpha \\ \Rightarrow & \left(1-x^{2}\right)\left(\frac{4-y^{2}}{4}\right)=\frac{4 \cos ^{2} \alpha+x^{2} y^{2}-4 x y \cos \alpha}{4} \\ \Rightarrow & \left(1-x^{2}\right)\left(4-y^{2}\right)=4 \cos ^{2} \alpha+x^{2} y^{2}-4 x y \cos \alpha \\ \Rightarrow & 4-4 x^{2}-y^{2}+x^{2} y^{2}=4 \cos ^{2} \alpha+x^{2} y^{2}-4 x y \cos \alpha \\ \Rightarrow & 4-4 x^{2}-y^{2}=4 \cos ^{2} \alpha-4 x y \cos \alpha \\ \Rightarrow & 4=4\left(1-\sin ^{2} \alpha\right)-4 x y \cos ^{2} \alpha+4 x^{2}+y^{2} \\ \Rightarrow & 4=4-4 \sin ^{2} \alpha-4 x y \cos \alpha+4 x^{2}+y^{2} \\ \Rightarrow & 4 x^{2}-4 x y \cos ^{2} \alpha+y^{2}=4 \sin ^{2} \alpha \end{aligned}$