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How do you evaluate$\arctan \left( { - \dfrac{{\sqrt 3 }}{3}} \right)$ ?

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint: Let us start solving this question by recalling that $\arctan \left( x \right) = {\tan ^{ - 1}}\left( x \right)$. Let us now assume that $\arctan \left( { - \dfrac{{\sqrt 3 }}{3}} \right)$ is equal to a variable and then we have to apply tangent on both sides. We already know that $\tan \left( {{{\tan }^{ - 1}}a} \right) = a$, let us proceed through the question by making use of the same. Next, we can also make use of the fact that if $\tan \theta = a$, for$a \in \mathbb{R}$, then the value of the angle $\theta $ lies in the interval $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ in order to get the required solution.

Complete step by step solution:
As per the question, we are supposed to find the value of $\arctan \left( { - \dfrac{{\sqrt 3 }}{3}} \right)$. We also know that $\arctan \left( x \right) = {\tan ^{ - 1}}\left( x \right)$.
Using the same we can get,
$ \arctan \left( { - \dfrac{{\sqrt 3 }}{3}} \right) = {\tan ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{3}} \right)$
Let us assume that ${\tan ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{3}} \right) = \alpha $---(1)
Now, let us apply tangents on both the left-hand and right-hand side of equation (1).
$ \Rightarrow \tan \left( {{{\tan }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{3}} \right)} \right) = \tan \alpha $---(2)
We know the identity that is, $\tan \left( {{{\tan }^{ - 1}}a} \right) = a$, for $a \in \mathbb{R}$. Now, let us use the same result in equation (2).
$ \Rightarrow - \dfrac{{\sqrt 3 }}{3} = \tan \alpha $----(3)
We also know that if $\tan \theta = a$, for$a \in \mathbb{R}$, then the value of angle $\theta $ lies in the interval $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$. So, we know that $\dfrac{{\sqrt 3 }}{3}$ can be simplified and can be written as $\dfrac{{\sqrt 3 }}{3} = \dfrac{1}{{\sqrt 3 }}$.
$\tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\sqrt 3 }}$, as the angle must lie in the interval $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$. Let us use this result in equation (3).
$ \Rightarrow \tan \left( { - \dfrac{\pi }{6}} \right) = \tan \alpha $---(4)
We know that if $\tan \theta = \tan a$, where $\theta $ belongs $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$, then the principal solution of angle $\alpha $ is equal to $\theta $. Let us use this identity in equation (4).
$ \therefore \alpha = - \dfrac{\pi }{6} = - {30^ \circ }$

Hence, the value of$\arctan \left( { - \dfrac{{\sqrt 3 }}{3}} \right) = - \dfrac{\pi }{6} = - {30^ \circ }$.

Note: Here, in this question we have assumed that we have to find the principal solution for the value of $\arctan \left( { - \dfrac{{\sqrt 3 }}{3}} \right)$. Before solving such a question, we first need to check whether the solution needs to be the principal solution or the general solution. The students should only report the angle that is present in the principal range of the inverse of tangent function.