Answer
$\tan(y)=\frac{\sqrt {1-x^{2}}}{x}$
Work Step by Step
Using Pythagorouses's theorem, the third side, $k$, of this right angled triangle can be found:
$1^{2}=x^{2}+ k^{2}$
$k^{2}=1-x^{2}$
$k=\sqrt {1-x^{2}}$ (since $k\gt0$
Using the triangle given in the diagram, the equation for $\tan(y)=\frac{opposite}{adjacent}$
opposite $=\sqrt {1-x^{2}}$, hypotenuse $=x$
Therefore $\tan(y)=\frac{\sqrt {1-x^{2}}}{x}$
