Question

If ${\cos }^{-1}x-{\cos }^{-1}\frac{y}{2}=\alpha$, then $4x^2-4xy\cos \alpha +y^2$ is equal to

• A. $2\sin 2\alpha$
• B. $4$
• C. $4{\sin }^2\alpha$
• D. $-4{\sin }^2\alpha$

Answer 1

Answer: (C)

$4{\sin }^2\alpha$

Given, ${\cos }^{-1}x-{\cos }^{-1}\frac{y}{2}=\alpha$

$\Rightarrow {\cos }^{-1}(\frac{xy}{2}+\sqrt{(1-x^2)(1-\frac{y^2}{4})})=\alpha$
$\Rightarrow {\cos }^{-1}\left(\frac{xy+\sqrt{4-y^2-4x^2+x^2{y}^2}}{2}\right)=\alpha$

$\Rightarrow 2\cos \alpha =xy+\sqrt{4-y^2-4x^2+x^2{y}^2}$

$\Rightarrow 4x^2+y^2-4xy\cos \alpha =4{\sin }^2\alpha$