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Question: How do you evaluate \({{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)\) ?...

How do you evaluate sec⁡−1(23){{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right) ?

Explanation

Solution

The term sec⁡−1{{\sec }^{-1}} in our expression is nothing but the inverse of sec⁡θ\sec \theta. So first we equate our expression with a variable and then convert sec⁡−1{{\sec }^{-1}} into sec⁡x\sec x by sending it to the other side of the equation. Since sec⁡θ\sec \theta is 1cos⁡θ\dfrac{1}{\cos \theta } , solving the given trigonometric quantity using cos⁡θ\cos \thetawould be easier. Find the value using the trigonometric values and then that θ\theta will be the solution.

Complete step by step solution:
The given trigonometric expression is, sec⁡−1(23){{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)
Let us first equate it to a variable.
Let us now consider x=sec⁡−1(23)x={{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)
Now apply sec⁡θ\sec \theta on both sides of the expression.
Then the expression will now be,
⇒sec⁡x=sec⁡(sec⁡−1(23))\Rightarrow \sec x=\sec \left( {{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right) \right)
Whenever there are a function and its inverse then,f(f−1(x))=xf\left( {{f}^{-1}}\left( x \right) \right)=x
On substituting the values in the above postulate, we get,
Hence our expression will be,
⇒sec⁡x=(23)\Rightarrow \sec x=\left( \dfrac{2}{\sqrt{3}} \right)
Since sec⁡θ\sec \theta is 1cos⁡θ\dfrac{1}{\cos \theta } converts out the expression in the same way.
⇒1cos⁡x=(23)\Rightarrow \dfrac{1}{\cos x}=\left( \dfrac{2}{\sqrt{3}} \right)
On inverting the entire equation, we get,
⇒cos⁡x=32\Rightarrow \cos x=\dfrac{\sqrt{3}}{2}
Using the trigonometric table for the values or using the right triangle and the Pythagoras theorem,
We know that cos⁡π6=32\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}
Therefore, the value of xx where the cos⁡\cos value is 32\dfrac{\sqrt{3}}{2} is when xx is equal to 30∘{{30}^{\circ }} .
Hence x=sec⁡−1(23)=30∘x={{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)={{30}^{\circ }} or x=π6x=\dfrac{\pi }{6}

Note: The inverse functions in trigonometry are also known as arc functions or anti trigonometric functions. They are majorly known as arc functions because they are most used to find the length of the arc needed to get the given or specified value. We can convert a function into an inverse function and vice versa. Always check when the trigonometric functions are given in degrees or radians. There is a lot of difference between both.1∘×π180=0.017Rad1{}^\circ \times \dfrac{\pi }{180}=0.017Rad. Express everything in sin⁡θ\sin \theta or cos⁡θ\cos \theta to easily evaluate.