Question

Find the principal value of ${\cot ^{ - 1}}\sqrt 3$.

Answer 1

Hint: The principal value of an inverse trigonometric function at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch. The principal value of the cotangent function is from $\left( {0,{{\pi }}} \right)$. From the table we will find the value of the function within the principal range.

Complete step-by-step answer:

The values of the cotangent functions are-

Function	$0^o$	$30^o$	$45^o$	$60^o$	$90^o$
cot	Not defined	$\sqrt 3$	1	$\dfrac{1}{{\sqrt 3 }}$	0

In the given question we need to find the principal value of ${\cot ^{ - 1}}\sqrt 3$. We know that for the cotangent function to be positive, the angle should be acute, that is less than $90^o$.

We know that the value of $\cot {30^{\text{o}}} = \sqrt 3$. So the value of the given expression can be calculated as-
${\cot ^{ - 1}}\left( {\sqrt 3 } \right) = {30^{\text{o}}}$
We know that ${{\pi }}$ rad = $180^o$, so
${\cot ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{{{\pi }}}{6}$
This is the required value.

Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. The principal value of cotangent in the question should always be between $0^o$ and $180^o$. This is because there can be infinite values of any inverse trigonometric functions. One common mistake is that students get confused between the tan and the cot functions, and often write the values of tan function or vice versa.