Question

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

Answer 1

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.