Question

Let ${\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{2}$ then prove that $\sqrt{1-x^2}\frac{dy}{dx}+\sqrt{1-y^2}=0$.

Answer 1

Given,

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

Now differentiating both sides with respect to $x$ we get,

$\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$

or, $\sqrt{1-x^2}\frac{dy}{dx}+\sqrt{1-y^2}=0$.