What is ${{\cot }^{-1}}\left( 3 \right)$ equal in decimal form?

Hint: We need to solve for the given expression and obtain the answer in degrees with a decimal. It can be done by first converting it in terms of tan and then check for the standard tan inverse values. If not found, then we can use the calculator to find the value of the expression accurately.

Complete step by step answer:
In order to solve this question, let us first explain what inverse means. The above expression in question can be represented as follows,
$\Rightarrow {{\cot }^{-1}}\left( 3 \right)=x$
We are equating it to x to show the inverse concept. Taking cot on both sides,
$\Rightarrow \cot \left( {{\cot }^{-1}}\left( 3 \right) \right)=\cot \left( x \right)$
For the term on the left-hand side, cot and cot inverse cancel giving us just 3.
$\Rightarrow 3=\cot \left( x \right)$
Hence, the inverse of a trigonometric term is supposed to give us an angle and here, x is an angle. We now convert this in terms of tan and this can be done by using the above equation itself.
Shifting both the terms to the denominator, the left-hand side term and right-hand side term,
$\Rightarrow \dfrac{1}{3}=\dfrac{1}{\cot \left( x \right)}$
We know that $\dfrac{1}{\cot \left( x \right)}$ can be written as $\tan \left( x \right),$
$\Rightarrow \dfrac{1}{3}=\tan \left( x \right)$
We refer to the standard angle values for tan and see that it is not a standard angle of the form $0{}^\circ ,30{}^\circ ,45{}^\circ ,60{}^\circ ,90{}^\circ ,etc.$ Hence, we use the calculator in degrees mode and apply the following equation to get x value,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{3} \right)=x$
We get the value of x as,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{3} \right)=x=18.4349{}^\circ $

Hence, the value of ${{\cot }^{-1}}\left( 3 \right)$ equal in decimal form is $18.4349{}^\circ .$

Note: We need to note that the inverse trigonometry is a very important concept for many mathematical problems. It gives us the angle values for the corresponding values of the function. We need to note that $\arctan \left( x \right)$ is same as ${{\tan }^{-1}}x$ and is not the same as $\dfrac{1}{\tan x}.$