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Question: How do you find the value of \({\sec ^{ - 1}}\left[ {\dfrac{{2\sqrt 3 }}{3}} \right]\)?...

How do you find the value of sec⁡−1[233]{\sec ^{ - 1}}\left[ {\dfrac{{2\sqrt 3 }}{3}} \right]?

Explanation

Solution

Here, in the given question, we need to find the value of sec⁡−1[233]{\sec ^{ - 1}}\left[ {\dfrac{{2\sqrt 3 }}{3}} \right]. As we can see here we are given an inverse trigonometric function. Inverse trigonometric functions are defined as the inverse functions of the basic trigonometric functions (trigonometric functions can be simply defined as functions of an angle of a triangle), they are also termed as arcus functions. As we know there is a relation between trigonometric function and an inverse trigonometric function which is given as; sec⁡−1(sec⁡x)=x{\sec ^{ - 1}}\left( {\sec x} \right) = x so, by using this formula we will find the value of sec⁡−1[233]{\sec ^{ - 1}}\left[ {\dfrac{{2\sqrt 3 }}{3}} \right] .

Formula used:
sec⁡−1x(sec⁡x)=x{\sec ^{ - 1}}x\left( {\sec x} \right) = x for all x∈(−∞,−1]∪[1,∞)x \in \left( { - \infty , - 1} \right] \cup \left[ {1,\infty } \right)

Complete step by step answer:
We have, sec⁡−1[233]{\sec ^{ - 1}}\left[ {\dfrac{{2\sqrt 3 }}{3}} \right]
As we know 33 can also be written as 3×3\sqrt 3 \times \sqrt 3 . Therefore, we will write the above written function as,
⇒sec⁡−1[233×3]\Rightarrow {\sec ^{ - 1}}\left[ {\dfrac{{2\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }}} \right]
On canceling out common terms, we get
⇒sec⁡−1[23]\Rightarrow {\sec ^{ - 1}}\left[ {\dfrac{2}{{\sqrt 3 }}} \right]
As we know sec⁡π6=23\sec \dfrac{\pi }{6} = \dfrac{2}{{\sqrt 3 }}. Therefore, we get
⇒sec⁡−1[sec⁡π6]\Rightarrow {\sec ^{ - 1}}\left[ {\sec \dfrac{\pi }{6}} \right]
As we know sec⁡−1(sec⁡x)=x{\sec ^{ - 1}}\left( {\sec x} \right) = x. Therefore, we get
⇒sec⁡−1[sec⁡π6]=π6\Rightarrow {\sec ^{ - 1}}\left[ {\sec \dfrac{\pi }{6}} \right] = \dfrac{\pi }{6}
Therefore, the value of sec⁡−1[233]{\sec ^{ - 1}}\left[ {\dfrac{{2\sqrt 3 }}{3}} \right] is π6\dfrac{\pi }{6}.

Note:
Remember that for any function ff and inverse of it i.e., f−1{f^{ - 1}}, f(f−1(x))=xf\left( {{f^{ - 1}}\left( x \right)} \right) = x and f−1(f(x))=x{f^{ - 1}}\left( {f\left( x \right)} \right) = x are same. Remember that inverse trigonometric functions do the opposite of the regular trigonometric functions. For example: inverse sec⁡\sec i.e., sec⁡−1{\sec ^{ - 1}} does the opposite of sec⁡\sec . The expression sec⁡−1x{\sec ^{ - 1}}x is not the same as 1secx\dfrac{1}{{secx}}. In other words, the −1 - 1 is not an exponent. Instead, it simply means inverse function. To solve this type of questions, one must remember the formulas of trigonometric functions and the interval in which they are defined.