JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 9)
Given that the inverse trigonometric function assumes principal values only. Let $$x, y$$ be any two real numbers in $$[-1,1]$$ such that $$\cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi$$.
Then, the minimum value of $$x^2+y^2+2 x y \sin \alpha$$ is
0
$$-$$1
$$\frac{1}{2}$$
$$\frac{-1}{2}$$
Explanation
$$\begin{aligned} & \cos ^{-1} x-\frac{\pi}{2}+\cos ^{-1} y=\alpha \\ & \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\ & \because \quad \alpha \in\left[\frac{-\pi}{2}, \pi\right] \\ & \text { then } \frac{\pi}{2}+\alpha \in\left(0, \frac{3 \pi}{2}\right) \\ & \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\frac{\pi}{2}+\alpha \\ & x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha \\ & x y+\sin \alpha=\sqrt{1-x^2} \sqrt{1-y^2} \\ & x^2 y^2+\sin ^2 \alpha+2 x y \sin \alpha=1-x^2-y^2+x^2 y^2 \\ & \underbrace{x^2+y^2+2 x y \sin \alpha}_E=\cos ^2 \alpha \end{aligned}$$
Now, minimum value of $$E$$ is 0.
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