If ${\tan ^{ - 1}}\left( {\sqrt 3 } \right) + {\cot ^{ - 1}}{\text{x}} = \dfrac{{{\pi }}}{2}$, then find the value of x.

Hint: The principal value of an inverse trigonometric function at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch. We have to find the principal value of $tan^{-1}(-1)$. We know that the principal value of $tan^{-1}(x)$ is given by $\left( { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right)$. The principal value of the cotangent function is from $\left( {0,{{\pi }}} \right)$. We will add the principal values of both the terms and then add them.

Complete step-by-step answer:
The values of the tangent functions are-



In the given question first we need to find the principal value of ${\tan ^{ - 1}}\left( {\sqrt 3 } \right)$. We know that for tangent function to be negative, the angle should be obtuse, that is greater than $90^o$. We know that the value of $\tan {60^o} = \sqrt 3 $ so
${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {60^{\text{o}}}$
We know that ${{\pi }}$ rad = $180^o$, so
${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{{{\pi }}}{3}$
So, substituting this value in the equation we get that-
${\cot ^{ - 1}}{\text{x}} = \dfrac{{{\pi }}}{2} - \dfrac{{{\pi }}}{3} = \dfrac{{{\pi }}}{6}$
${\text{x}} = \cot \dfrac{{{\pi }}}{6} = \sqrt 3 $
This is the required value of x.

Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. The range of principal values of tangent function is from $-90^o$ to $90^o$, and for cotangent function is from $0^o$ to $180^o$. This is because there can be infinite values of any inverse trigonometric functions. We can also directly use the identity that-
$\begin{align}
  &{\tan ^{ - 1}}{\text{x}} + {\cot ^{ - 1}}{\text{x}} = \dfrac{{{\pi }}}{2} \\
  &{\cot ^{ - 1}}{\text{x}} = \dfrac{{{\pi }}}{2} - {\tan ^{ - 1}}\sqrt 3 \\
  &{\cot ^{ - 1}}{\text{x}} = {\cot ^{ - 1}}\sqrt 3 \\
  &{\text{x}} = \sqrt 3 \\
\end{align} $