Simplify Inverse Trigonometry Expressions Help! I need to simplify this expression. I'm not even sure where to start. 
$$
\tan{(\arccos{(\frac{x}{4})})}
$$

Let  $\displaystyle y=\tan\left(\cos^{-1}\left(\frac{x}{4}\right)\right)\;,$ Now Put $\displaystyle \cos^{-1}\left(\frac{x}{4}\right)=\phi\;,$
Then $\displaystyle \frac{x}{4} = \cos \phi\;,$ Then $\displaystyle \sec \phi = \frac{4}{x},$ So using $\displaystyle \tan \phi = \pm \sqrt{\sec^2 \phi -1} = \pm \frac{\sqrt{16-x^2}}{x}$
So We get $\displaystyle y = \tan \phi = \pm \frac{\sqrt{16-x^2}}{x}$
