Solve the differential equation: $\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} } $ $$\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} }$$
Simplified it to: $$ \int \frac{1}{\sqrt{1-y^2}} dy = \int \frac{1}{\sqrt{1-x^2}} dx$$
$$\implies\sin^{-1} y = ( \sin^{-1} x + C)$$
$$y = (\sin (\sin^{-1} x) \cos C_1) + \cos(\sin^{-1} x) (\sin C_1)$$ 
How to I simplify $\cos(\sin^{-1} x)$ to get the final answer ?

You can simplify it like this
$$ y = \sin(\sin^{-1} x + C) = \cos (C) \sin(\sin^{-1}x) + \sin (C)\cos(\sin^{-1}x) \\
= \cos (C)x + \sin(C)\sqrt{1-x^2} $$
The explanation is
$$ \cos(\sin^{-1}x) = \sqrt{1-\sin^2(\sin^{-1}x)} = \sqrt{1-x^2} $$
