Answer
$\dfrac{\sqrt {1-x^2}}{x}$
Work Step by Step
Suppose $\theta =\cos^{-1} x $
This gives: $ x =\cos \theta $
Since, $ r=\sqrt {a^2+y^2} \implies a=\sqrt {r^2-y^2}$
This implies that $ a=\sqrt {1^2-x^2}=\sqrt {1-x^2}$
Therefore, we have $\tan(\cos^{-1} x)=\dfrac{\sqrt {1-x^2}}{x}$
