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

How do you evaluatearctan⁡(−33)\arctan \left( { - \dfrac{{\sqrt 3 }}{3}} \right) ?

Explanation

Solution

Let us start solving this question by recalling that arctan⁡(x)=tan⁡−1(x)\arctan \left( x \right) = {\tan ^{ - 1}}\left( x \right). Let us now assume that arctan⁡(−33)\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⁡(tan⁡−1a)=a\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⁡θ=a\tan \theta = a, fora∈Ra \in \mathbb{R}, then the value of the angle θ\theta lies in the interval (−π2,π2)\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⁡(−33)\arctan \left( { - \dfrac{{\sqrt 3 }}{3}} \right). We also know that arctan⁡(x)=tan⁡−1(x)\arctan \left( x \right) = {\tan ^{ - 1}}\left( x \right).
Using the same we can get,
arctan⁡(−33)=tan⁡−1(−33)\arctan \left( { - \dfrac{{\sqrt 3 }}{3}} \right) = {\tan ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{3}} \right)
Let us assume that tan⁡−1(−33)=α{\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).
⇒tan⁡(tan⁡−1(−33))=tan⁡α\Rightarrow \tan \left( {{{\tan }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{3}} \right)} \right) = \tan \alpha---(2)
We know the identity that is, tan⁡(tan⁡−1a)=a\tan \left( {{{\tan }^{ - 1}}a} \right) = a, for a∈Ra \in \mathbb{R}. Now, let us use the same result in equation (2).
⇒−33=tan⁡α\Rightarrow - \dfrac{{\sqrt 3 }}{3} = \tan \alpha----(3)
We also know that if tan⁡θ=a\tan \theta = a, fora∈Ra \in \mathbb{R}, then the value of angle θ\theta lies in the interval (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right). So, we know that 33\dfrac{{\sqrt 3 }}{3} can be simplified and can be written as 33=13\dfrac{{\sqrt 3 }}{3} = \dfrac{1}{{\sqrt 3 }}.
tan⁡(π6)=13\tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\sqrt 3 }}, as the angle must lie in the interval (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right). Let us use this result in equation (3).
⇒tan⁡(−π6)=tan⁡α\Rightarrow \tan \left( { - \dfrac{\pi }{6}} \right) = \tan \alpha---(4)
We know that if tan⁡θ=tan⁡a\tan \theta = \tan a, where θ\theta belongs (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right), then the principal solution of angle α\alpha is equal to θ\theta . Let us use this identity in equation (4).
∴α=−π6=−30∘\therefore \alpha = - \dfrac{\pi }{6} = - {30^ \circ }

Hence, the value ofarctan⁡(−33)=−π6=−30∘\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⁡(−33)\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.