Answer
$tan(arccos\frac{x}{3})=\frac{\sqrt {9-x^2}}{x}$
Work Step by Step
Let $u=arccos\frac{x}{3}~~$ (Range: $0\lt u\lt\pi$)
Then: $\frac{x}{3}=cos~u$
$sin^2u+cos^2u=1$
$sin^2u=1-cos^2u$
$sin^2u=1-(\frac{x}{3})^2=1-\frac{x^2}{9}=\frac{9-x^2}{9}$
$sin~u=\frac{\sqrt {9-x^2}}{3}~~$ ($0\lt u\lt\pi$)
$tan(arccos\frac{x}{3})=tan~u=\frac{sin~u}{cos~u}=\frac{\frac{\sqrt {9-x^2}}{3}}{\frac{x}{3}}=\frac{\sqrt {9-x^2}}{x}$
