<s>[INST] $y=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$ and $y$ can be expanded in ascending powers of $x$ If $y=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$ and assuming that $y$ can be expanded in ascending powers of $x$ in the form $a_0+a_1x+a_2x^2+....+a_nx^n$... prove that $(n+1)a_{n+1}=na_{n-1}$
How to proceed in this question? Is it possible to write $y=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$ in ascending powers of x?
 [/INST] HINT:
$$y\sqrt{1-x^2}=\sin^{-1}x$$
Differentiating both sides wrt $x$
$$y_1\sqrt{1-x^2}-y\cdot\dfrac x{\sqrt{1-x^2}}=\dfrac1{\sqrt{1-x^2}}$$ 
$$\iff(1-x^2)y_1-xy=1$$
As $y=\sum_{r=0}^\infty a_rx^r, y_1=\sum_{r=1}^\infty ra_rx^{r-1}$
Compare the coefficients of $x^{r+2}$
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