What is $\cot \left( -{{120}^{\circ }} \right)$ as a function of a positive acute angle?

Hint: Use the conversion $\cot x=\dfrac{\cos x}{\sin x}$ to simplify. Further use the properties of the sine and cosine functions given as $\cos \left( -\theta \right)=\cos \theta $ and $\sin \left( -\theta \right)=-\sin \theta $ to make the angle positive. To convert the 120 degrees into the acute angle write it as ${{120}^{\circ }}=\left( {{180}^{\circ }}-{{60}^{\circ }} \right)$ and use the properties of the $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta $ and $\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta $ by considering the fact that the cosine function is negative in the second quadrant while sine being positive.

Complete step-by-step solution:
Here we have been provided with the expression $\cot \left( -{{120}^{\circ }} \right)$ and we are asked to write it in a form such that the angle becomes positive and acute.
Now, Cotangent function is the ratio of cosine and the sine function also can be said as the inverse of the tangent function. Mathematically it is given as:
$\begin{align}
  & \Rightarrow \cot x=\dfrac{\cos x}{\sin x} \\
 & \Rightarrow \cot \left( -{{120}^{\circ }} \right)=\dfrac{\cos \left( -{{120}^{\circ }} \right)}{\sin \left( -{{120}^{\circ }} \right)} \\
\end{align}$
Using the properties of sine and cosine function given as $\cos \left( -\theta \right)=\cos \theta $ and $\sin \left( -\theta \right)=-\sin \theta $ we get,
$\begin{align}
  & \Rightarrow \cot \left( -{{120}^{\circ }} \right)=\dfrac{\cos \left( {{120}^{\circ }} \right)}{-\sin \left( {{120}^{\circ }} \right)} \\
 & \Rightarrow \cot \left( -{{120}^{\circ }} \right)=-\left( \dfrac{\cos \left( {{120}^{\circ }} \right)}{\sin \left( {{120}^{\circ }} \right)} \right) \\
\end{align}$
Here we can see that 120 degrees is an obtuse angle and we need to make it acute, i.e. and angle between 0 and 90 degrees. So we can write 120 degrees as ${{120}^{\circ }}=\left( {{180}^{\circ }}-{{60}^{\circ }} \right)$, therefore the expression becomes: -
$\Rightarrow \cot \left( -{{120}^{\circ }} \right)=-\left( \dfrac{\cos \left( {{180}^{\circ }}-{{60}^{\circ }} \right)}{\sin \left( {{180}^{\circ }}-{{60}^{\circ }} \right)} \right)$
Now, $\left( {{180}^{\circ }}-{{60}^{\circ }} \right)$ lies in the second quadrant and we know that cosine function is negative in the second quadrant while sine function is positive, so using the formulas $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta $ and $\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta $ we get,
$\begin{align}
  & \Rightarrow \cot \left( -{{120}^{\circ }} \right)=\dfrac{\cos \left( {{60}^{\circ }} \right)}{\sin \left( {{60}^{\circ }} \right)} \\
 & \Rightarrow \cot \left( -{{120}^{\circ }} \right)=\cot \left( {{60}^{\circ }} \right) \\
\end{align}$
Hence, the above relation is our answer.

Note: You must remember the sign of all six trigonometric functions in all four quadrants. The above properties that are used in the solution are due to the periodicity of trigonometric functions. Note that you can also write 120 degrees as $\left( {{90}^{\circ }}+{{30}^{\circ }} \right)$ but in this case the function gets changed. Sine becomes cosine and cosine becomes sine so the answer will turn out to be $\tan \left( {{30}^{\circ }} \right)$ which can be converted into the cotangent function by using the complementary angle rule given as $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \left( \theta \right)$..