Q108 / 225 · JEE Main 2005 · 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
    $2 \sin 2 \alpha$
  • B
    4
  • Correct answer
    C
    $4 \sin ^2 \alpha$
  • D
    $-4 \sin ^2 \alpha$

Solution

$$ \begin{aligned} & \cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha \\ & \cos ^{-1}\left(\frac{x y}{2}+\sqrt{\left(-x^2\right)\left(1-\frac{y^2}{4}\right)}\right)=\alpha \\ & \cos ^{-1}\left(\frac{x y+\sqrt{4-y^2-4 x^2+x^2 y^2}}{2}\right)=\alpha \\ & \Rightarrow 4-y^2-4 x^2+x^2 y^2=4 \cos ^2 \alpha+x^2 y^2-4 x y \cos \alpha \\ & \Rightarrow 4 x^2+y^2-4 x y \cos \alpha=4 \sin ^2 \alpha \end{aligned} $$

Hence, the correct answer is option: 'C'.

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