Question

Let $(x,y)$ be such that
${\sin }^{-1}\left(ax\right)+{\cos }^{-1}\left(y\right)+{\cos }^{-1}$ $\left(bxy\right)$ $=\frac{\pi }{2}$.
Match the statements in Column I with the statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4\times 4$ matrix given in the $ORS$.

Answer 1

(A) If $a=1,b=0$ then ${\sin }^{-1}x+{\cos }^{-1}y=0$

$\Rightarrow {\sin }^{-1}x=-{\cos }^{-1}y$

$\Rightarrow x^2+y^2=1$.

(B) If $a=1$ and $b=1$, then

${\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}xy=\frac{\pi }{2}$

$\Rightarrow {\cos }^{-1}x-{\cos }^{-1}y={\cos }^{-1}xy$

$\Rightarrow xy+\sqrt{1-x^2}\sqrt{1-y^2}=$ $xy$ (taking cosine on both the sides)

$\Rightarrow (x^2-1)(y^2-1)=0$

(C) If $a=1,b=2$

$\Rightarrow {\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}\left(2xy\right)=\frac{\pi }{2}$

$\Rightarrow xy+\sqrt{1-x^2}\sqrt{1-y^2}=2xy$

$\Rightarrow x^2+y^2=1$ (on squaring).

(D) If $a=2$ and $b=2$ then

${\sin }^{-1}\left(2x\right)+{\cos }^{-1}\left(y\right)+{\cos }^{-1}\left(2xy\right)=\frac{\pi }{2}$

$\Rightarrow 2xy+\sqrt{1-4x^2}\sqrt{1-y^2}=2xy$

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