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(B) We are given $\cos ^{-1} x - \sin ^{-1} y = \alpha$. Using the identity $\sin ^{-1} y = \frac{\pi}{2} - \cos ^{-1} y$,we get: $\cos ^{-1} x - (\fr

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$0$

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(B) We are given $\cos ^{-1} x - \sin ^{-1} y = \alpha$. Using the identity $\sin ^{-1} y = \frac{\pi}{2} - \cos ^{-1} y$,we get: $\cos ^{-1} x - (\frac{\pi}{2} - \cos ^{-1} y) = \alpha$ $\cos ^{-1} x + \cos ^{-1} y = \frac{\pi}{2} + \alpha$. Since $\alpha \in [-\frac{\pi}{2}, \pi]$,we have $\frac{\pi}{2} + \alpha \in [0, \frac{3\pi}{2}]$. Taking cosine on both sides: $\cos(\cos ^{-1} x + \cos ^{-1} y) = \cos(\frac{\pi}{2} + \alpha)$ $xy - \sqrt{1-x^2}\sqrt{1-y^2} = -\sin \alpha$ $xy + \sin \alpha = \sqrt{1-x^2}\sqrt{1-y^2}$. Squaring both sides: $(xy + \sin \alpha)^2 = (1-x^2)(1-y^2)$ $x^2y^2 + 2xy \sin \alpha + \sin^2 \alpha = 1 - x^2 - y^2 + x^2y^2$ $x^2 + y^2 + 2xy \sin \alpha = 1 - \sin^2 \alpha$ $x^2 + y^2 + 2xy \sin \alpha = \cos^2 \alpha$. The minimum value of $\cos^2 \alpha$ is $0$,which occurs when $\alpha = \frac{\pi}{2}$. Thus,the minimum value is $0$.

Column $I$	Column $II$
$(A)$ If $a=1$ and $b=0$,then $(x, y)$	$(p)$ lies on the circle $x^2+y^2=1$
$(B)$ If $a=1$ and $b=1$,then $(x, y)$	$(q)$ lies on $(x^2-1)(y^2-1)=0$
$(C)$ If $a=1$ and $b=2$,then $(x, y)$	$(r)$ lies on $y=x$
$(D)$ If $a=2$ and $b=2$,then $(x, y)$	$(s)$ lies on $(4x^2-1)(y^2-1)=0$

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

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