Q. The differential equation for which $\sin^{- 1}x + \sin^{- 1}y = c$ is given by
A) $\sqrt{1 - x^{2}}dy + \sqrt{1 - y^{2}}dx = 0$
B) $\sqrt{1 - x^{2}}\ dx + \sqrt{1 - y^{2}}\ dy = 0$
C) $\sqrt{1 - x^{2}}\ dx - \sqrt{1 - y^{2}}\ dy = 0$
D) $\sqrt{1 - x^{2}}\ dy - \sqrt{1 - y^{2}}\ dx = 0$
Explanation
(a)
Given, $\sin^{- 1}x + \sin^{- 1}y = c$
$\Rightarrow$ $\frac{dx}{\sqrt{1 - x^{2}}} + \frac{dy}{\sqrt{1 - y^{2}}} = 0$
$\Rightarrow$ $\sqrt{1 - y^{2}}\ dx + \sqrt{1 - x^{2}}\ dy = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.