Answer
The identity is verified.
$arcsin~x=arctan\frac{x}{\sqrt {1-x^2}}$
Work Step by Step
Let $y=arcsin~x~~$
Then: $x=sin~y~~$
$sin^2y+cos^2y=1$
$cos^2y=1-sin^2y=1-x^2$
$\frac{1}{cos^2y}=\frac{1}{1-x^2}$
$sec^2~y=\frac{1}{1-x^2}$
$sec^2~y-1=\frac{1}{1-x^2}-1=\frac{1}{1-x^2}-\frac{1-x^2}{1-x^2}=\frac{x^2}{1-x^2}$
$tan^2y=\frac{x^2}{1-x^2}$
$tan~y=\frac{x}{\sqrt {1-x^2}}$
$y=arctan\frac{x}{\sqrt {1-x^2}}=arcsin~x$
