Given the inverse trigonometric function assumes principal values… - Vidyadip

MCQ

Given the inverse trigonometric function assumes principal values only. Let $\mathrm{x}, \mathrm{y}$ be any two real numbers in $[-1,1]$ such that $\cos ^{-1} \mathrm{x}-\sin ^{-1} \mathrm{y}=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi \text {. }$ Then, the minimum value of $x^2+y^2+2 x y \sin \alpha$ is

A
$-1$
✓
$0$
C
$\frac{-1}{2}$
D
$\frac{1}{2}$

✓

Answer

Correct option: B.

$0$

b
$ \cos ^{-1} x-\left(\frac{\pi}{2}-\cos ^{-1} y\right)=\alpha $

$ \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha $

$ \alpha \in\left[-\frac{\pi}{2}, \pi\right], \frac{\pi}{2}+\alpha \in\left[0, \frac{3 \pi}{2}\right] $

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

$ x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha $

$ (x y+\sin \alpha)=\left(1-x^2\right)\left(1-y^2\right) $

$ x^2 y^2+2 x y \operatorname{sina}+\sin ^2 a=1-x^2-y^2+x^2 y^2 $

$ x^2+y^2+2 x y \sin \alpha=1-\sin ^2 \alpha $

$ x^2+y^2+2 x y \sin \alpha=\cos ^2 \alpha$

Min. value of $\cos ^2 \alpha=0$

At $\alpha=\frac{\pi}{2}$

Option ($2$) is correct

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