Find the Exact Value tan(arccos(x))

Draw a triangle in the plane with vertices ${\displaystyle (x,\sqrt{1^2-x^2})}$, ${\displaystyle (x,0)}$, and the origin. Then ${\displaystyle \arccos \left(x\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (x,\sqrt{1^2-x^2})}$. Therefore, ${\displaystyle \tan \left(\arccos \left(x\right)\right)}$ is ${\displaystyle \frac{\sqrt{1-x^2}}{x}}$.