Find the Exact Value tan(arccos(w))

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