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

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

Explanation

Solution

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⁡(−7π3)\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(g(x))f\left( {g\left( x \right)} \right), composition function.
And we know in the composition functions f(g(x))f\left( {g\left( x \right)} \right), we will firstly solve the function g(x)g\left( x \right) and then put the obtained value in ff function.
So, here g(x)g\left( x \right) in (1) is sec⁡(−7π3)\sec \left( { - \dfrac{{7\pi }}{3}} \right),
Therefore, we will firstly proceed with sec⁡(−7π3)−−−−−(2)\sec \left( { - \dfrac{{7\pi }}{3}} \right)-----(2)
Now we know that, sec⁡(−x)=sec⁡x\sec \left( { - x} \right) = \sec x
So, we can write
sec⁡(−7π3)=sec⁡(7π3)\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( {\dfrac{{7\pi }}{3}} \right)
We can also write 7π3=2π+π3\dfrac{{7\pi }}{3} = 2\pi + \dfrac{\pi }{3}, so we get
sec⁡(−7π3)=sec⁡(2π+π3)\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( {2\pi + \dfrac{\pi }{3}} \right)
Now we know that sec⁡(2π+θ)=sec⁡θ\sec \left( {2\pi + \theta } \right) = \sec \theta , which is valid for any every trigonometric function,
So, we get
sec⁡(−7π3)=sec⁡(π3)−−−−−(3)\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( {\dfrac{\pi }{3}} \right)-----(3)
Now substituting value of sec⁡(−7π3)\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⁡(−7π3)\sec \left( { - \dfrac{{7\pi }}{3}} \right), and we know that sec⁡\sec is positive in fourth quadrant then we can write that
sec⁡(−7π3)=sec⁡(2π−7π3)\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( {2\pi - \dfrac{{7\pi }}{3}} \right)
sec⁡(−7π3)=sec⁡(−π3)\sec \left( { - \dfrac{{7\pi }}{3}} \right) = \sec \left( { - \dfrac{\pi }{3}} \right)
Now we know that sec⁡(−x)=sec⁡x\sec \left( { - x} \right) = \sec x, so using this we can write that
sec⁡(−7π3)=sec⁡(π3)\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⁡(−7π3)\sec \left( { - \dfrac{{7\pi }}{3}} \right) in (1), and proceed similarly.