Question

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

Answer 1

Hint: Here, in the given question, we need to find the value of ${\sec ^{ - 1}}\left[ {\dfrac{{2\sqrt 3 }}{3}} \right]$. 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; ${\sec ^{ - 1}}\left( {\sec x} \right) = x$ so, by using this formula we will find the value of ${\sec ^{ - 1}}\left[ {\dfrac{{2\sqrt 3 }}{3}} \right]$ .

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

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

Note:
Remember that for any function $f$ and inverse of it i.e., ${f^{ - 1}}$, $f\left( {{f^{ - 1}}\left( x \right)} \right) = x$ and ${f^{ - 1}}\left( {f\left( x \right)} \right) = x$ are same. Remember that inverse trigonometric functions do the opposite of the regular trigonometric functions. For example: inverse $\sec $ i.e., ${\sec ^{ - 1}}$ does the opposite of $\sec $. The expression ${\sec ^{ - 1}}x$ is not the same as $\dfrac{1}{{secx}}$. In other words, the $ - 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.