Let (x, y) be such that $sin^{-1}(ax) + cos^{-1}y - cos^{-1} (bxy) = \frac{\pi}{2}$ If a = 1 and b = 2 , then (x, y) lies on

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

$x^2 + y^2 = 1$

If a = 1, b = 2, then $sin^{-1}(ax) + cos^{-1}y - cos^{-1} (bxy) = \frac{\pi}{2}$ $⇒ sin^{-1}x + cos^{-1}y - cos^{-1} (2xy) = \frac{\pi}{2}$ $⇒ cos^{-1} x - cos^{-1} y = cos^{-1}(2xy)$ $⇒  xy + \sqrt{1-x^2} \sqrt{1-y^2} = 2xy $ $⇒ x^2 + y^2 = 1 $