Question

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

Answer 1

Hint: 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 ${\cot ^{ - 1}}( - \sqrt 3 )$. The solution in which the absolute value of the angle is the least is called the principal solution and for example, the value of \[\cos {0^ \circ }\]is 1 so as of $\cos 2\pi ,4\pi ,..$ is also 1. But 0 is known as the principal value.

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