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Question: The principal value of \({\cot ^{ - 1}}\left( { - 1} \right)\) is:- A.\(\dfrac{\pi }{4}\) B.\[ -...

The principal value of cot⁡−1(−1){\cot ^{ - 1}}\left( { - 1} \right) is:-
A.π4\dfrac{\pi }{4}
B.−π4 - \dfrac{\pi }{4}
C.3π4\dfrac{{3\pi }}{4}
D.None of these

Explanation

Solution

We will first write cot⁡−1(−1)=x{\cot ^{ - 1}}\left( { - 1} \right) = x. Then, find the angle where cot⁡x=1\cot x = 1. Now, as we know that cot can be negative in the second and fourth quadrant only, find their corresponding values in those quadrants. The values in the cartesian plane between 0⩽x<2π0 \leqslant x < 2\pi such that cot⁡x=−1\cot x = - 1 will be the principal values of cot⁡−1(−1)=x{\cot ^{ - 1}}\left( { - 1} \right) = x.

Complete step-by-step answer:
We have to find the principal value of cot⁡−1(−1){\cot ^{ - 1}}\left( { - 1} \right). That is, we want to find the angle where cot⁡x=−1\cot x = - 1 in the cartesian plane such that 0⩽x<2π0 \leqslant x < 2\pi
We will know let cot⁡−1(−1)=x{\cot ^{ - 1}}\left( { - 1} \right) = x, which implies that cot⁡x=−1\cot x = - 1
Since, the value is negative, and cot of any angle is negative in the second and fourth quadrant only.
Hence, the value of xx is in the second and fourth quadrant.
We also know that cot⁡(π4)=1\cot \left( {\dfrac{\pi }{4}} \right) = 1
Therefore, we will have the corresponding angle in the second and fourth quadrant.
Let us first find the value in the second quadrant.
For, writing the angle in second quadrant whose value will be negative of cot⁡(π4)\cot \left( {\dfrac{\pi }{4}} \right) , we will subtract π4\dfrac{\pi }{4} from π\pi .
Hence, we have
cot⁡(π−π4)=−1 cot⁡(3π4)=−1  \cot \left( {\pi - \dfrac{\pi }{4}} \right) = - 1 \\\ \cot \left( {\dfrac{{3\pi }}{4}} \right) = - 1 \\\
Hence, the principal value is 3π4\dfrac{{3\pi }}{4} in the second quadrant.
Now, we will find the value in the fourth quadrant.
For, writing the angle in second quadrant whose value will be negative of cot⁡(π4)\cot \left( {\dfrac{\pi }{4}} \right) , we will subtract π4\dfrac{\pi }{4} from 2π2\pi .
Hence, we have
cot⁡(2π−π4)=−1 ⇒cot⁡(7π4)=−1  \cot \left( {2\pi - \dfrac{\pi }{4}} \right) = - 1 \\\ \Rightarrow \cot \left( {\dfrac{{7\pi }}{4}} \right) = - 1 \\\
Hence, the principal value is 7π4\dfrac{{7\pi }}{4} in the fourth quadrant.
Therefore, option C is correct.

Note: A cartesian plane is divided into four quadrants. All trigonometry ratios are positive in the first quadrant, in the second quadrant only sin and cosec are positive, in third quadrant only ratios which are positive are of tan and cot, in fourth quadrant cos and secant will give positive answers.