Question

Find the value of ${\sec ^{ - 1}}\left\{ {\sec \left( { - \dfrac{{7\pi }}{3}} \right)} \right\}$.

Answer 1

Hint:In the above question we have to find the value of the trigonometric function, in order to solve this question we will use the trigonometric formula ${\sec ^{ - 1}}\left\{ {\sec \left( \theta \right)} \right\} = \theta ,\theta \in \left[ {0,\pi } \right] - \left\{ {\dfrac{\pi }{2}} \right\}$, by firstly reducing $\sec \left( { - \dfrac{{7\pi }}{3}} \right)$ in such a way that its angle satisfy the above formula.

Complete step-by-step answer:
In the above question we have to find the value of
${\sec ^{ - 1}}\left\{ {\sec \left( { - \dfrac{{7\pi }}{3}} \right)} \right\} -----(1)$
In the above trigonometric function, it is of the form $f\left( {g\left( x \right)} \right)$, composition function.
And we know in the composition functions $f\left( {g\left( x \right)} \right)$, we will firstly solve the function $g\left( x \right)$ and then put the obtained value in $f$ function.
So, here $g\left( x \right)$ in (1) is $\sec \left( { - \dfrac{{7\pi }}{3}} \right)$,
Therefore, we will firstly proceed with $\sec \left( { - \dfrac{{7\pi }}{3}} \right)-----(2)$
Now we know that, $\sec \left( { - x} \right) = \sec x$
So, we can write
\[\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( {\dfrac{{7\pi }}{3}} \right)\]
We can also write $\dfrac{{7\pi }}{3} = 2\pi + \dfrac{\pi }{3}$, so we get
\[\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( {2\pi + \dfrac{\pi }{3}} \right)\]
Now we know that $\sec \left( {2\pi + \theta } \right) = \sec \theta $, which is valid for any every trigonometric function,
So, we get
$\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( {\dfrac{\pi }{3}} \right)-----(3)$
Now substituting value of $\sec \left( { - \dfrac{{7\pi }}{3}} \right)$ from (3) in (1), we get
${\sec ^{ - 1}}\left\{ {\sec \left( { - \dfrac{{7\pi }}{3}} \right)} \right\} = {\sec ^{ - 1}}\left\{ {\sec \left( {\dfrac{\pi }{3}} \right)} \right\}-----(4)$
Since we know that ${\sec ^{ - 1}}\left\{ {\sec \left( \theta \right)} \right\} = \theta ,\theta \in \left[ {0,\pi } \right] - \left\{ {\dfrac{\pi }{2}} \right\}-----(5)$
Now in (4), $\theta = \dfrac{\pi }{3},{\text{ where }}\dfrac{\pi }{3} \in \left[ {0,\pi } \right] - \left\{ {\dfrac{\pi }{2}} \right\}$
So, using (5) in (4), we get
${\sec ^{ - 1}}\left\{ {\sec \left( { - \dfrac{{7\pi }}{3}} \right)} \right\} = \dfrac{\pi }{3}$
So, this is the required answer.

Note:The alternative way to do this question is-
We will consider $\sec \left( { - \dfrac{{7\pi }}{3}} \right)$, and we know that $\sec $ is positive in fourth quadrant then we can write that
$\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( {2\pi - \dfrac{{7\pi }}{3}} \right)$
$\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( { - \dfrac{\pi }{3}} \right)$
Now we know that $\sec \left( { - x} \right) = \sec x$, so using this we can write that
 $\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( {\dfrac{\pi }{3}} \right)$
Hence the above equation (3) ,where we have reached in another way.
Hence after obtaining (3), we can put the value of $\sec \left( { - \dfrac{{7\pi }}{3}} \right)$ in (1), and proceed similarly.