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
Answer & explanation
Correct answer: option 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 $