If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$, where $-1 \leq x \leq 1,-2 \leq y \leq 2, x \leq…
If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$, where $-1 \leq x \leq 1,-2 \leq y \leq 2, x \leq \frac{y}{2}$, then for all $x, y$. $4 x^2-4 x y \cos a+y^2$ is equal to
$2 \sin ^2 a$
$4 \sin ^2 a$
$4 \cos ^2 a+2 x^2 y^2$
$4 \sin ^2 a-2 x^2 y^2$
Solution
$\begin{aligned} & \cos ^{-1} x-\cos ^{-1} \frac{y}{2}=a \\ & \Rightarrow \cos ^{-1}\left(x \cdot \frac{y}{2}+\sqrt{1-x^2} \sqrt{1-\frac{y^2}{4}}\right)=a \\ & \Rightarrow \frac{x y}{2}+\frac{\sqrt{\left(1-x^2\right)\left(4-y^2\right)}}{2}=\cos a \\ & \Rightarrow\left(1-x^2\right)\left(4-y^2\right)=2 \cos a-x y \\ & \Rightarrow 4-y^2-4 x^2+x^2 y^2=4 \cos ^2 a+x^2 y^2-4 \cos a \cdot x y \\ & \Rightarrow 4-4 \cos ^2 a=4 x^2-4 x y \cos ^2 a+y^2 \\ & \Rightarrow 4 x^2-4 x y \cos a+y^2=4 \sin ^2 a\end{aligned}$