Answer

Verified

373.2k+ views

**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.

Recently Updated Pages

In the equation x2+y2+2gx+2fy+c0 represents a circle class 10 maths CBSE

How do you Factor completely 2x3 + 5x2 37x 60 class 10 maths CBSE

How do you prove the identity dfrac11 cos x + dfrac11 class 10 maths CBSE

How do you solve 2x2 5x 12 0 by completing the squ class 10 maths CBSE

How do you solve 2x+46 class 10 maths CBSE

How do you simplify dfracm42m4 and write it using only class 10 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail

Name 10 Living and Non living things class 9 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths