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