Question

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

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

Answer 1

Answer: (B)

Key Idea : $cos^{-1}\,x\pm cos^{-1} y=cos^{-1}\left(xy\mp\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right).$
We have, $cos^{-1}\,x-cos^{-1} \frac{y}{2}=\alpha$
$\Rightarrow cos^{-1}\left(\frac{xy}{2}+\sqrt{1-x^{2}}\sqrt{1-\frac{y^{2}}{4}}\right)=\alpha$
$\Rightarrow \frac{xy}{2}+\sqrt{1-x^{2}}\sqrt{1-\frac{y^{2}}{4}}=cos\,\alpha$
$\Rightarrow 2\sqrt{1-x^{2}}\sqrt{1-\frac{y^{2}}{4}}=2\,cos\,\alpha-xy$
$\Rightarrow 4-4x^{2}-y^{2}+x^{2}y^{2}=4\,cos^{2}\,\alpha+x^{2}y^{2}-2xy\,cos\,\alpha$
$\Rightarrow 4x^{2}-4\,xy\,cos\,\alpha+y^{2}=4\,sin^{2}\,\alpha.$