Find the Exact Value tan(arccos(u))

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