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The solution of the differential equation $\frac{dy}{dx}+\sqrt{\frac{1-y^2}{1-x^2}}=0,$ is |
$tan^{-1} x + cot^{-1} x = C$ $sin^{-1} x + sin^{-1} y = C$ $sec^{-1}x+cosec^{-1}x=C$ none of thes |
$sin^{-1} x + sin^{-1} y = C$ |
The correct answer is option (2) : $sin^{-1} x + sin^{-1} y = C$ We have, $\frac{dy}{dx}+\sqrt{\frac{1-y^2}{1-x^2}}=0$ $⇒\frac{1}{\sqrt{1-y^2}}dy+\frac{1}{\sqrt{1-x^2}}dx=0$ On integrating, wehave $sin^{-1} x + sin^{-1} y = C$ |
