How do you evaluate the expression $ \cot \left( -\pi \right) $ ?

Hint: We explain the process of finding values for associated angles. We find the rotation and the position of the angle for $ \left( -\pi \right) $ . We explain the changes that are required for that angle. Depending on those things we find the solution.

Complete step-by-step answer:
We need to find the ratio value for $ \cot \left( -\pi \right) $ .
For general form of $ \cot \left( x \right) $ , we need to convert the value of x into the closest multiple of $ \dfrac{\pi }{2} $ and add or subtract a certain value $ \alpha $ from that multiple of $ \dfrac{\pi }{2} $ to make it equal to $ x $ .
Let’s assume $ x=k\times \dfrac{\pi }{2}+\alpha $ , $ k\in \mathbb{Z} $ . Here we took addition of $ \alpha $ . We also need to remember that $ \left| \alpha \right|\le \dfrac{\pi }{2} $ .
Now we take the value of k. If it’s even then keep the ratio as cot and if it’s odd then the ratio changes to cot ratio from tan.
Then we find the position of the given angle as quadrant value measured in counter clockwise movement from the origin and the positive side of X-axis.
If the angel falls in the first or third quadrant then the sign remains positive but if it falls in the second or fourth quadrant then the sign becomes negative.
Depending on the sign and ratio change the final angle becomes $ \alpha $ from x.
As the given angle is negative the movement procedure which is counter clockwise, actually becomes clockwise. Now we won’t consider the sign any more.
For the given angle $ \pi $ , we can express it as
 $ \pi =2\times \dfrac{\pi }{2}+0 $ .
The value of k is even which means the trigonometric ratio remains
 $ \cot \left( x \right) $ .
The position of the angle is in the second quadrant. The angle completes the half-circle 1 times.
Therefore, the sign becomes negative.
The final form becomes
 $ \cot \left( -\pi \right)=\cot \left( 2\times \dfrac{\pi }{2}+0 \right)=-\cot \left( 0 \right)=\dfrac{1}{0} $ .
Therefore, the value of $ \cot \left( -\pi \right) $ is undefined.
So, the correct answer is “undefined”.

Note: We need to remember that the easiest way to avoid the change of ratio thing is to form the multiple of $ \pi $ instead of $ \dfrac{\pi }{2} $ . It makes the multiplied number always even. In that case we don’t have to change the ratio. If $ x=k\times \pi +\alpha =2k\times \dfrac{\pi }{2}+\alpha $ . Value of $ 2k $ is always even.