Question

How do you calculate $ \arctan (\dfrac{{ - \sqrt 3 }}{3}) $ ?

Answer 1

Hint: We use trigonometry to find the relation between two sides and an angle of a right-angled triangle. The trigonometric functions are given as the ratio of two sides of the right angles triangle. In the given question, we have to evaluate $ \arctan (\dfrac{{ - \sqrt 3 }}{3}) $ , $ \arctan A $ means the tangent inverse of the value whose tangent is $ A $ . Thus, we have to find the tangent inverse of the angle whose tangent is $ \dfrac{{ - \sqrt 3 }}{3} $ , we know that $ A = \tan \theta = \dfrac{{ - \sqrt 3 }}{3} $ , using trigonometric identities and formulas, we can find out the correct answer

Complete step-by-step answer:
We are given that $ \tan \theta = \dfrac{{ - \sqrt 3 }}{3} $
 And we have to find $ \arctan (\dfrac{{ - \sqrt 3 }}{3}) $ , that is, $ {\tan ^{ - 1}}(\dfrac{{ - \sqrt 3 }}{3}) $
From the definition of $ \arctan (\dfrac{{ - \sqrt 3 }}{3}) $ , we know that $ \tan \theta = \dfrac{{ - \sqrt 3 }}{3} $
 $ \Rightarrow \tan \theta = - \dfrac{1}{{\sqrt 3 }} $
We know that $ \tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }} $
Now, the tangent of an angle is negative in the second and the fourth quadrant,
So
 $
  \tan \theta = \tan (\pi - \dfrac{\pi }{6}) = \tan (2\pi - \dfrac{\pi }{6}) \\
   \Rightarrow \tan \theta = \tan \dfrac{{5\pi }}{6} = \tan \dfrac{{11\pi }}{6} \\
   \Rightarrow \theta = \dfrac{{5\pi }}{6} = \dfrac{{11\pi }}{6} \;
  $
Thus the general value of $ \theta $ is $ \theta = n\pi - (\dfrac{\pi }{6}),n \in \mathbb{Z} $
Hence $ \arctan (\dfrac{{ - \sqrt 3 }}{3}) $ is $ \theta = n\pi - (\dfrac{\pi }{6}),n \in \mathbb{Z} $
So, the correct answer is “ $ \theta = n\pi - (\dfrac{\pi }{6}),n \in \mathbb{Z} $ ”.

Note: Before solving these types of questions, we should ensure that we have to find the general solution or the principal solution. In this question, the answer cannot lie between the principal branch that is $ (\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}) $ , so we find out the general solution. The graph is divided into four quadrants and different trigonometric functions have different signs in different quadrants. The sign of tangent function is negative only in the second and fourth quadrant, an angle lying in the second quadrant is smaller than $ \pi $ and greater than $ \dfrac{\pi }{2} $ ,and an angle lying in the fourth quadrant is smaller than $ 2\pi $ and greater than $ \dfrac{{3\pi }}{2} $ that’s why we subtract the angle from $ \pi $ and $ 2\pi $ .