Answer
$$\arcsin x = \arctan \left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right),{\text{ }}\left| x \right| < 1$$
Work Step by Step
$$\eqalign{
& {\text{*From the triangle }} \cr
& {\text{tan}}\theta = \frac{x}{{\sqrt {1 - {x^2}} }}{\text{ for }} - 1 < x < 1 \cr
& \arctan \left( {{\text{tan}}\theta } \right) = \arctan \left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right) \cr
& \theta = \arctan \left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right) \cr
& {\text{*From the triangle}} \cr
& x = {\text{sin}}\theta \cr
& \arcsin x = \arcsin \left( {\sin \theta } \right) \cr
& \arcsin x = \theta \cr
& {\text{Substituting }}\theta = \arctan \left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right) \cr
& \arcsin x = \arctan \left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right),{\text{ for }} - 1 < x < 1{\text{ or}} \cr
& \arcsin x = \arctan \left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right),{\text{ }}\left| x \right| < 1 \cr} $$
