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 ?

Put,

$y$ = $\sin \alpha$  and  $x$ = $\sin \beta$   

calculate $dy$ and $dx$,

$dy$ = $\cos \alpha$$.d \alpha$ and $dx$ = $\cos \beta$$.d\beta$.  

The equation will be something like,  

=> $\int \cot\alpha $ $d\alpha$= $\int \cot\beta$ $d\beta$
  => $\log$ |$sin$ $\alpha$| + $c_1$ = $\log$ |$sin$ $\beta$| + $c_2$
  => $\left(\frac{sin \alpha} {sin\beta}\right)$ = $e^{c_2 - c_1}$ = $c$
  => $sin$ $\alpha$ = $sin$ $\beta$ $.c$

Now, replace $sin$ $\alpha$ and $sin$ $\beta$, with $y$ and $x$.  

$y$ = $x$ $.c$

