Question

Find the principal value of ${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)$.

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 sec function is $\left[ {0,{{\pi }}} \right] - \left\{ {\dfrac{{{\pi }}}{2}} \right\}$. We will find the value of the given function inside the given principal range to get our answer.

Complete step-by-step answer:
The principal values of the trigonometric functions are-

Function	$0^o$	$30^o$	$45^o$	$60^o$	$90^o$
sec	1	$\dfrac{2}{{\sqrt 3 }}$	$\sqrt 2 $	2	Not defined

In the given question we need to find the principal value of ${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)$. We know that for the secant function to be positive, the angle should be acute, that is less than $90^o$.
We know that the value of $\sec {30^{\text{o}}} = \dfrac{2}{{\sqrt 3 }}$. So the value of the given expression can be calculated as-
${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right) = {30^{\text{o}}}$
We know that ${{\pi }}$ rad = $180^o$, so
${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right) = \dfrac{{{\pi }}}{6}$
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 secant in the question should always be from $0^{o}$ to $180^{o}$, excluding $90^{o}$. This is because there can be infinite values of any inverse trigonometric functions. A common mistake is that the students use $60^o$ instead of $30^o$ or vice versa, in order to solve the problem quickly.