Match the following List-I with List-II.

	List-I		List-II
(A)	If ${\left({\sin }^{-1}x\right)}^2+{\left({\sin }^{-1}y\right)}^2=\frac{{\pi }^2}{2}.$ then $x^3+y^3=$	(p)	0
(B)	${\left({\cos }^{-1}x\right)}^2+{\left({\cos }^{-1}y\right)}^2=2{\pi }^2,$  then $x^5+y^5$ can be	(q)	-2
(C)	${\left({\sin }^{-1}x\right)}^2\times {\left({\cos }^{-1}y\right)}^2=\frac{{\pi }^4}{4},$  then $x-y$ can be	(r)	2
(D)	$|{\sin }^{-1}x-{\sin }^{-1}y|=\pi ,$ then $x-y$ can be	(s)	-1

 

A.  ${\left({\sin }^{-1}x\right)}^2+{\left({\sin }^{-1}y\right)}^2=\frac{{\pi }^2}{2}$
 $\Rightarrow {\left({\sin }^{-1}x\right)}^2={\left({\sin }^{-1}y\right)}^2=\frac{{\pi }^2}{4}$
 $\Rightarrow {\sin }^{-1}x=\pm \frac{\pi }{2}.{\sin }^{-1}y=\pm \frac{\pi }{2}$
 $\Rightarrow x=\pm 1andy=\pm 1$
 $\Rightarrow x^3+y^3=-2,0,2$
B.  ${\left({\cos }^{-1}x\right)}^2+{\left({\cos }^{-1}y\right)}^2=2{\pi }^2$
 $\Rightarrow {\left({\cos }^{-1}x\right)}^2={\left({\cos }^{-1}y\right)}^2={\pi }^2$
 $\Rightarrow x=y=1$
 $\Rightarrow x^5+x^5=-2$
C.  ${\left({\sin }^{-1}x\right)}^2{\left({\cos }^{-1}y\right)}^2=\frac{{\pi }^4}{4}$
$\Rightarrow {\left({\sin }^{-1}x\right)}^2=\frac{{\pi }^4}{4}$  and  ${\left({\cos }^{-1}y\right)}^2={\pi }^2$
 $\Rightarrow \left({\sin }^{-1}x\right)=\pm \frac{\pi }{2}\left({\cos }^{-1}y\right)=\pi$
 $\Rightarrow x=\pm 1andy=-1$
 $x-y=0.2$
D.  $|{\sin }^{-1}x-{\sin }^{-1}y|=\pi$
 $\Rightarrow {\sin }^{-1}x=-\frac{\pi }{2}$ and  ${\sin }^{-1}y=\frac{\pi }{2}$ 
 $\Rightarrow x=-1andy=-1$
$\Rightarrow x-y=-2,2$