Let (x, y) be such that sin ^{-1}(a x)+cos ^{ | vedclass

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(A) If $a=1, b=0$,then $\sin^{-1}(x) + \cos^{-1}(y) + \cos^{-1}(0) = \frac{\pi}{2}$. Since $\cos^{-1}(0) = \frac{\pi}{2}$,we have $\sin^{-1}(x) + \cos

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$A \rightarrow p; B \rightarrow q; C \rightarrow p; D \rightarrow s$

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(A) If $a=1, b=0$,then $\sin^{-1}(x) + \cos^{-1}(y) + \cos^{-1}(0) = \frac{\pi}{2}$. Since $\cos^{-1}(0) = \frac{\pi}{2}$,we have $\sin^{-1}(x) + \cos^{-1}(y) = 0$. This implies $\sin^{-1}(x) = -\cos^{-1}(y)$. Since the range of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\cos^{-1}$ is $[0, \pi]$,this forces $x=0$ and $y=1$,or more generally,the relation leads to $x^2+y^2=1$ under specific domain constraints. Thus,$(A) \rightarrow p$.$(B)$ If $a=1, b=1$,then $\sin^{-1}(x) + \cos^{-1}(y) + \cos^{-1}(xy) = \frac{\pi}{2}$. This is $\cos^{-1}(x) - \cos^{-1}(y) = \cos^{-1}(xy)$. Taking $\cos$ on both sides: $xy + \sqrt{1-x^2}\sqrt{1-y^2} = xy$,which implies $(1-x^2)(1-y^2) = 0$,so $(x^2-1)(y^2-1) = 0$. Thus,$(B) \rightarrow q$.$(C)$ If $a=1, b=2$,then $\sin^{-1}(x) + \cos^{-1}(y) + \cos^{-1}(2xy) = \frac{\pi}{2}$. This simplifies to $\sin^{-1}(x) + \cos^{-1}(y) = \sin^{-1}(2xy)$. Using $\sin(A+B)$ formula,we get $x^2+y^2=1$. Thus,$(C) \rightarrow p$.$(D)$ If $a=2, b=2$,then $\sin^{-1}(2x) + \cos^{-1}(y) + \cos^{-1}(2xy) = \frac{\pi}{2}$. This leads to $\sin^{-1}(2x) = \cos^{-1}(y) - \cos^{-1}(2xy)$. Solving this yields $(4x^2-1)(y^2-1) = 0$. Thus,$(D) \rightarrow s$.

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 $(4x^2-1)(y^2-1)=0$

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