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If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$,
then $4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to
Options:
then $4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to
Solution:
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Verified Answer
The correct answer is:
$4 \sin ^{2} \alpha$
$\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$
$\Rightarrow \cos ^{-1}\left(\frac{x y}{2}+\sqrt{\left(1-x^{2}\right)\left(1-\frac{y^{2}}{4}\right)}\right)=\alpha$
$\Rightarrow \cos ^{-1}\left(\frac{x y+\sqrt{4-y^{2}-4 x^{2}+x^{2} y^{2}}}{2}\right)=\alpha$
$\Rightarrow 4 x^{2}+y^{2}-4 x y \cos \alpha=4 \sin ^{2} \alpha$
$\Rightarrow \cos ^{-1}\left(\frac{x y}{2}+\sqrt{\left(1-x^{2}\right)\left(1-\frac{y^{2}}{4}\right)}\right)=\alpha$
$\Rightarrow \cos ^{-1}\left(\frac{x y+\sqrt{4-y^{2}-4 x^{2}+x^{2} y^{2}}}{2}\right)=\alpha$
$\Rightarrow 4 x^{2}+y^{2}-4 x y \cos \alpha=4 \sin ^{2} \alpha$
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