$\ y = \dfrac {\sin^{-1}x}{\sqrt{1-x^2}} \cdots(Given)$

$y {\sqrt{1-x^2}}= \sin^{-1}x$

• $Squarring \ both\ sides,\ we\ get$

$y^2 ({1-x^2})=( {\sin^{-1}x})^2 $

• $Now,\ we\ differentiate\ w.r.\ t.\ x$

$= y^2 . (-2x) + (1-x^2).2yy_1= 2 \sin^{-1}x. \dfrac {1} {\sqrt{1-x^2}} \cdots(1) $

• $Substitute \dfrac {\sin^{-1}x} {\sqrt{1-x^2}}=y\ \cdots in\ (1)$

$-2xy^2 + (1-x^2).2yy_1= 2y$

• $Dividing\ by\ '2y'\ we\ get,$

$ -xy + (1-x^2).y_1= 1$

$(1-x^2).y_1= 1 + xy$

• $Applying\ Leibnitz\ theorem, we\ get$

$(1-x^2).y_{n+1}+n.(-2x)y_n +\dfrac{n(n-1)} {2!}.(-2)y_{n-1}= 0 + [x.y_n+n.1.y_{n-1}]$

$(1-x^2)y_{n+1}-2nxy_n - {n(n-1)}y_{n-1}= xy_n+ny_{n-1}]$

$(1-x^2)y_{n+1}-2nxy_n + (-n^2+n)y_{n-1}- xy_n-ny_{n-1}=0$

$(1-x^2)y_{n+1}+ (-2nx - x)y_n + (-n^2+n -n)y_{n-1}=0$

$(1-x^2)y_{n+1}- (2n + 1)xy_n -n^2y_{n-1}=0 \cdots Hence\ Proved.$