Inverse Circular Functions 3 Question 2
2. 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
(a) $2 \sin ^{2} \alpha$
(2019 Main, 10 April II)
(b) $4 \cos ^{2} \alpha+2 x^{2} y^{2}$
(c) $4 \sin ^{2} \alpha$
(d) $4 \sin ^{2} \alpha-2 x^{2} y^{2}$
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Answer:
Correct Answer: 2. (c)
Solution:
- Given equation is
$$ \begin{aligned} & \cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha \text {, where }-1 \leq x \leq 1, \\ & \quad-2 \leq y \leq 2 \text { and } x \leq \frac{y}{2} \\ & \cos ^{-1} x \frac{y}{2}+\sqrt{1-x^{2}} \sqrt{1-(y / 2)^{2}}=\alpha \\ & {\left[\because \cos ^{-1} x-\cos ^{-1} y=\cos ^{-1}\left(x y+\sqrt{1-x^{2}} \sqrt{\left.1-y^{2}\right)},\right.\right.} \\ & \Rightarrow \frac{x y}{2}+\sqrt{1-x^{2}} \sqrt{1-(y / 2)^{2}}=\cos \alpha \\ & \Rightarrow \quad \sqrt{1-x^{2}} \sqrt{1-(y / 2)^{2}}=\cos \alpha-\frac{x y}{2} \end{aligned} $$
On squaring both sides, we get
$$ \left(1-x^{2}\right) 1-\frac{y^{2}}{4}=\cos ^{2} \alpha+\frac{x^{2} y^{2}}{4}-2 \frac{x y}{2} \cos \alpha $$
$\Rightarrow 1-x^{2}-\frac{y^{2}}{4}+\frac{x^{2} y^{2}}{4}=\cos ^{2} \alpha+\frac{x^{2} y^{2}}{4}-x y \cos \alpha$
$\Rightarrow x^{2}+\frac{y^{2}}{4}-x y \cos \alpha=1-\cos ^{2} \alpha$
$\Rightarrow 4 x^{2}-4 x y \cos \alpha+y^{2}=4 \sin ^{2} \alpha$
