## Calculus (3rd Edition)

$\frac{\sqrt{5}}{2}$
Assume that $\cos\theta =\frac{2}{3}$, then by solving the triangle, we have: $$\tan \theta=\frac{\sqrt{5}}{2}.$$ Now, since $\theta =\cos^{-1}\frac{2}{3}$, we have $$\tan(\cos^{-1}\frac{2}{3})=\tan\theta=\frac{\sqrt{5}}{2}.$$