Given that, $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha \quad \Rightarrow \quad \cos ^{-1}\left(\frac{x y}{2}+\sqrt{1-x^2} \sqrt{1-\frac{y^2}{4}}\right)=\alpha$
$\Rightarrow \quad \frac{x y}{2}+\sqrt{1-x^2} \sqrt{1-\frac{y^2}{4}}=\cos \alpha \quad \Rightarrow \quad 2 \sqrt{1-x^2} \sqrt{1-\frac{y^2}{4}}=2 \cos \alpha-x y$
On squaring both sides, we get (दोनों पक्षों का वर्ग करने पर)
$\frac{4\left(1-x^2\right)\left(4-y^2\right)}{4}=4 \cos ^2 \alpha+x^2 y^2-4 x y \cos \alpha \Rightarrow 4-4 x^2-y^2+x^2 y^2=4 \cos ^2 \alpha+x^2 y^2-4 x y \cos \alpha \\$
$\Rightarrow \quad 4 x^2-4 x y \cos \alpha+y^2=4 \sin ^2 \alpha$
Hence Option 3 is Correct.
