# if

Question:

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 $x, y, 4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to: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 $x, y, 4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to:

1. (1) $4 \sin ^{2} \alpha$

2. (2) $2 \sin ^{2} \alpha$

3. (3) $4 \sin ^{2} \alpha-2 x^{2} y^{2}$

4. (d) $4 \cos ^{2} \alpha+2 x^{2} y^{2}$

Correct Option: 1

Solution:

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

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

$\Rightarrow \frac{x y}{2}+\frac{\sqrt{1-x^{2}} \sqrt{4-y^{2}}}{2}=\cos \theta$

$\Rightarrow x y+\sqrt{1-x^{2}} \sqrt{4-y^{2}}=2 \cos \alpha$

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

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

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