If $\cos ^{-1} \mathrm{x}-\cos ^{-1} \frac{\mathrm{y}}{2}=\alpha$
where $-1 \leq \mathrm{x} \leq 1,-2 \leq \mathrm{y} \leq 2, \mathrm{x} \leq \frac{\mathrm{y}}{2}$
then for all $x, y, 4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to
Correct Option: , 3
$\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$
$\cos \left(\cos ^{-1} x-\cos ^{-1} \frac{y}{2}\right)=\cos \alpha$
$\Rightarrow x \times \frac{y}{2}+\sqrt{1-x^{2}} \sqrt{1-\frac{y^{2}}{4}}=\cos \alpha$
$\Rightarrow\left(\cos \alpha-\frac{x y}{2}\right)^{2}=\left(1-x^{2}\right)\left(1-\frac{y^{2}}{4}\right)$
$x^{2}+\frac{y^{2}}{4}-x y \cos \alpha=1-\cos ^{2} \alpha=\sin ^{2} \alpha$
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