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Question: The principal value of \({\cot ^{ - 1}}( - \sqrt 3 )\) is: A.\(\dfrac{{ - \pi }}{6}\) B.\(\dfrac...

The principal value of cot⁡−1(−3){\cot ^{ - 1}}( - \sqrt 3 ) is:
A.−π6\dfrac{{ - \pi }}{6}
B.π6\dfrac{\pi }{6}
C.7π6\dfrac{{7\pi }}{6}
D.5π6\dfrac{{5\pi }}{6}

Explanation

Solution

We will convert the given equation into the standard form i.e., in terms of cot and then by using its range, we will determine the principal value of cot⁡−1(−3){\cot ^{ - 1}}( - \sqrt 3 ). The solution in which the absolute value of the angle is the least is called the principal solution and for example, the value of cos⁡0∘\cos {0^ \circ }is 1 so as of cos⁡2π,4π,..\cos 2\pi ,4\pi ,.. is also 1. But 0 is known as the principal value.

Complete step-by-step answer:
Here, we are given the function cot⁡−1(−3){\cot ^{ - 1}}( - \sqrt 3 ) and we are required to find its principal value.
We know the inverse trigonometric identity cot⁡−1(−x)=π−cot⁡−1(x);x∈R{\cot ^{ - 1}}\left( { - x} \right) = \pi - {\cot ^{ - 1}}\left( x \right);x \in {\mathbf{R}}
Supposing that the value of x = 3\sqrt 3 in the above equation, we can say that 3\sqrt 3 ∈R \in {\mathbf{R}}.
∴cot⁡−1−3=π−cot⁡−13\therefore {\cot ^{ - 1}}\sqrt { - 3} = \pi - {\cot ^{ - 1}}\sqrt 3
Let y=cot⁡−1(3)y = {\cot ^{ - 1}}(\sqrt 3 )
⇒cot⁡y=3\Rightarrow \cot y = \sqrt 3
⇒cot⁡y=3=cot⁡π6\Rightarrow \cot y = \sqrt 3 = \cot \dfrac{\pi }{6}
Therefore, we get y=π6y = \dfrac{\pi }{6}
We know that the range of the principal value of cot⁡−1θ{\cot ^{ - 1}}\theta is (0,π)\left( {0,\pi } \right).
For positive values of the given function, we have the principal value θ\theta and for negative values of the given function, we will have the principal value as π−θ\pi - \theta . Here,−3\sqrt { - 3} is negative for the function, hence we will use the principal value π−θ\pi - \theta here to make the principal value lie in the range of the given function.
Therefore, the principal value of the given function cot⁡−1(−3){\cot ^{ - 1}}( - \sqrt 3 ) is
⇒π−π6=5π6\Rightarrow \pi - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6}
Hence, the principal value will be 5π6\dfrac{{5\pi }}{6}.
Option(D) is correct.

Note: In such questions where you are asked the principal value of any particular function, we generally get confused with which value is the principal value. Like if the solution is a negative value in the any case where we were asked the principal value of cot⁡−1(x){\cot ^{ - 1}}(x)(say), then we can’t say that the principal value will be the solution we reach after solving but the actual principal value would be π−x\pi - x.