Question

Let (x, y) be such that $si{n}^{-1}\left(ax\right)+co{s}^{-1}\left(y\right)+co{s}^{-1}\left(bxy\right)=\pi /2$

Answer 1

As ${\sin }^{-1}\left(ax\right)+{\cos }^{-1}y+{\cos }^{-1}\left(bxy\right)=\frac{\pi }{2}$
A) For $a=1,b=0$
${\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}0=\frac{\pi }{2}\Rightarrow {\sin }^{-1}x+{\cos }^{-1}y=0\Rightarrow {\sin }^{-1}x=-{\cos }^{-1}y=-{\sin }^{-1}\sqrt{1-y^2}\Rightarrow x=-\sqrt{1-y^2}\Rightarrow x^2=1-y^2\Rightarrow x^2+y^2=1$
B) For $a=1,b=1$
${\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}xy=\frac{\pi }{2}\Rightarrow {\cos }^{-1}xy=\frac{\pi }{2}-{\sin }^{-1}x-{\cos }^{-1}y={\cos }^{-1}x-{\cos }^{-1}y\Rightarrow {\cos }^{-1}xy={\cos }^{-1}(xy+\sqrt{1-x^2}\sqrt{1-y^2})\Rightarrow xy=xy+\sqrt{1-x^2}\sqrt{1-y^2}\Rightarrow (1-x^2)(1-y^2)=0$
C) For $a=1,b=2$
${\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}2xy=\frac{\pi }{2}\Rightarrow co{s}^{-1}\left(2xy\right)=co{s}^{-1}(xy+\sqrt{1-x^2}\sqrt{1-y^2})\Rightarrow xy=\sqrt{1-x^2}\sqrt{1-y^2}\Rightarrow x^2{y}^2=1-x^2-y^2+x^2{y}^2\Rightarrow x^2+y^2=1$
D) For $a=2,b=2$
${\sin }^{-1}2x+{\cos }^{-1}y+{\cos }^{-1}2xy=\frac{\pi }{2}\Rightarrow co{s}^{-1}\left(2xy\right)=si{n}^{-1}y-si{n}^{-1}\left(2x\right)\Rightarrow co{s}^{-1}\left(2xy\right)=co{s}^{-1}(\sqrt{1-y^2}\sqrt{1-4x^2}+2xy)\Rightarrow \sqrt{1-y^2}\sqrt{1-4x^2}=0\Rightarrow (4x^2-1)(y^2-1)=0$