How do you simplify \[\tan \left( {{{\cos }^{ - 1}}x} \right)\]?

Hint: To solve this question first we will assume $\cos^{-1}x= \theta$ . We use standard identity \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] to get the value of $\sin {\theta}$. Then using these values we will calculate the required value.

Complete step by step answer:
The given question is \[\tan \left( {{{\cos }^{ - 1}}x} \right)\]
Take an assumption that, \[{\cos ^{ - 1}}x = \theta \]
Which implies that \[\cos \theta = x\]
We all know that \[{\cos ^2}\theta + {\sin ^2}\theta = 1\]
So, from here, we can write \[\sin \theta = \pm \sqrt {1 - {{\cos }^2}x} \]
\[ \Rightarrow \sin \theta = \pm \sqrt {1 - {x^2}} \]
So, the expression becomes \[\tan \theta \]
We all know that \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
So, we can write \[\tan \theta = \pm \dfrac{{\sqrt {1 - {x^2}} }}{x}\]
Finally, we can conclude that, \[\tan \left( {{{\cos }^{ - 1}}x} \right) = \pm \dfrac{{\sqrt {1 - {x^2}} }}{x}\]

Note:
Remember the identity \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] which is a standard identity in trigonometry. The result we got consists of both positive and negative values. We have to consider both the values.
\[{\sin ^{ - 1}}x\] and \[{\cos ^{ - 1}}x\] are defined only when \[ - 1 \leqslant x \leqslant 1\].
And also remember the identity \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\].