Question

How do you evaluate $\tan \left( -{{30}^{\circ }} \right)$?

Answer 1

Hint: To find the value of $\tan \left( -{{30}^{\circ }} \right)$, we have to determine the quadrant to which the given angle belongs to. Since we are given a negative angle, this means that it is in the fourth quadrant. We know that the tangent function is negative in the fourth quadrant and is positive in the first quadrant. So the value of $\tan \left( -{{30}^{\circ }} \right)$ will be equal to the negative of its value in the first quadrant. So we have to shift the given angle in the first quadrant, calculate its tangent and take its negative to finally get the required value.

Complete step-by-step answer:
In the above question, we are asked to calculate the value of $\tan \left( -{{30}^{\circ }} \right)$. Let us represent this value in the below equation as
$y=\tan \left( -{{30}^{\circ }} \right).......(i)$
Since the angle of $-{{30}^{\circ }}$ is negative, this means that it is taken in the clockwise direction with reference to the x-axis. This means that this angle belongs to the fourth quadrant. We know that the tangent function is negative in the fourth quadrant. So it must be equal to the negative of its value in the first quadrant. For this, we shift the given angle to the first quadrant. So we need to take the angle in the opposite direction, that is, the anti-clockwise direction. Therefore after shifting the angle into the first quadrant, we obtain the value of the angle as ${{30}^{\circ }}$. So the above equation can also be written as
\[y=-\tan \left( {{30}^{\circ }} \right)\]
Now, we know that $\tan \left( {{30}^{\circ }} \right)=\dfrac{1}{\sqrt{3}}$. Substituting this in the above equation, we get
$y=-\dfrac{1}{\sqrt{3}}........(ii)$
Finally, on equating the equations (i) and (ii), we get
$\tan \left( -{{30}^{\circ }} \right)=-\dfrac{1}{\sqrt{3}}$
Hence, the value of $\tan \left( -{{30}^{\circ }} \right)$ is equal to $-\dfrac{1}{\sqrt{3}}$.

Note: The function $\tan x$ is an odd function. So we could also use the identity $\tan \left( -x \right)=-\tan x$ for solving the above question. We are asked to evaluate the value of $\tan \left( -{{30}^{\circ }} \right)$ in the above question. So on putting $x={{30}^{\circ }}$ in the identity $\tan \left( -x \right)=-\tan x$ we will get the same value as obtained in the above solution.