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