Q52 / 125 · BITSAT 2015 · Mathematics
If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$,
then $4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to
- A$2 \sin 2 \alpha$
- B4
- Correct answerC$4 \sin ^{2} \alpha$
- D$-4 \sin ^{2} \alpha$
Solution
$\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$
$\Rightarrow \cos ^{-1}\left(\frac{x y}{2}+\sqrt{\left(1-x^{2}\right)\left(1-\frac{y^{2}}{4}\right)}\right)=\alpha$
$\Rightarrow \cos ^{-1}\left(\frac{x y+\sqrt{4-y^{2}-4 x^{2}+x^{2} y^{2}}}{2}\right)=\alpha$
$\Rightarrow 4 x^{2}+y^{2}-4 x y \cos \alpha=4 \sin ^{2} \alpha$
Hence, the correct answer is option: 'C'.