Question

How do you find the exact value of $\tan \left[ {{{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right]$?

Answer 1

Hint: In this question, we are given a trigonometric expression and we have been asked to find the exact value. Find the angle at which the value of sin is $\dfrac{{ - 1}}{2}$ . Then, put the angle besides tan and find the value of tan at that angle. This value will be your given answer.

Complete step-by-step solution:
We are given a trigonometric expression and we have to find its value.
$ \Rightarrow \tan \left[ {{{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right]$ …. (given)
We know that the value of $\sin $ is equal to $\dfrac{1}{2}$ at$\dfrac{\pi }{6}$ .
But when is the value of $\sin $ equal to$ - \dfrac{1}{2}$? We know that $\sin $ is negative in \[{3^{rd}}\] and \[{4^{th}}\] quadrant.
So, let’s consider the \[{4^{th}}\] quadrant.
How to navigate to the \[{4^{th}}\] quadrant? There are two ways:
1) We can enter the \[{4^{th}}\] quadrant from \[{3^{rd}}\] quadrant. It is done as follows:
$ \Rightarrow \dfrac{{3\pi }}{2} + x$
2) We can also enter the \[{4^{th}}\] quadrant from the \[{1^{st}}\] quadrant. It is done as follows:
$ \Rightarrow - \dfrac{\pi }{6}$
I will use the \[{2^{nd}}\] method.
Hence, ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{ - \pi }}{6}$
Putting in the equation,
$ \Rightarrow \tan \left[ {\dfrac{{ - \pi }}{6}} \right]$
We can write it as:
$ \Rightarrow - \tan \left[ {\dfrac{\pi }{6}} \right]$
Putting the value of $\tan \dfrac{\pi }{6}$,
$ \Rightarrow - \dfrac{1}{{\sqrt 3 }}$

Hence, $\tan \left[ {{{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right] = \dfrac{{ - 1}}{{\sqrt 3 }}$

Note: As I said above that the value of sin is negative in the $3^{rd}$ quadrant also, so, why did I not use the $3^{rd}$ quadrant? It can be explained as below:
$ \Rightarrow \tan \left[ {{{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right]$ …. (given)
We know that $\sin \left( { - \theta } \right) = - \sin \theta $ . Using this in the above equation,
$ \Rightarrow \tan \left[ { - {{\sin }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right]$
Now, it has taken the form of $\tan \left( { - \theta } \right)$. We know that $\tan \left( { - \theta } \right) = - \tan \theta $ .
$ \Rightarrow - \tan \left[ {{{\sin }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right]$
Now, tan is negative in \[{2^{nd}}\] and \[{4^{th}}\] quadrant. And sin is negative in \[{3^{rd}}\] and \[{4^{th}}\] quadrant. The common quadrant between the two is the \[{4^{th}}\] quadrant. Hence, that is why we chose the \[{4^{th}}\] quadrant.