Question

How do you solve $\arctan \left( { - \dfrac{1}{{\sqrt 3 }}} \right)$?

Answer 1

In order to determine the value of the above question, first we will assume it be as $\theta$, and as we know that the range of the inverse function of tangent is in the interval $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right),$so$\theta \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$. Using the fact that $\tan \left( { - \theta } \right) = - \tan \theta$ and $\tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\sqrt 3 }}$, we obtain the $\theta$ to be $- \dfrac{\pi }{6}$.

Complete step by step solution:
We are given a trigonometric expression $\arctan \left( { - \dfrac{1}{{\sqrt 3 }}} \right)$
$\Rightarrow \arctan \left( { - \dfrac{1}{{\sqrt 3 }}} \right)$
Let’s assume $\arctan \left( { - \dfrac{1}{{\sqrt 3 }}} \right) = \theta$
As we know that the range of inverse of tangent is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.
Therefore $\theta \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$
Note that: $\tan \left( { - \theta } \right) = - \tan \theta$
As we know the value of $\tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\sqrt 3 }}$
So, $\tan \left( { - \dfrac{\pi }{6}} \right) = - \tan \left( {\dfrac{\pi }{6}} \right) = - \dfrac{1}{{\sqrt 3 }}$,where $\left( { - \dfrac{\pi }{6}} \right) \in \left( { - \dfrac{\pi }{2},0} \right)$
We can write,
$\tan \left( { - \dfrac{\pi }{6}} \right) = - \dfrac{1}{{\sqrt 3 }}$
Taking inverse of tangent on both the sides,we get
$\Rightarrow {\tan ^{ - 1}}\left( {\tan \left( { - \dfrac{\pi }{6}} \right)} \right) = {\tan ^{ - 1}}\left( { \- \dfrac{1}{{\sqrt 3 }}} \right) \\\ \Rightarrow - \dfrac{\pi }{6} = {\tan ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 3 }}} \right)or \\\ \Rightarrow {\tan ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 3 }}} \right) = - \dfrac{\pi }{6} \\\$
Therefore, the value of $\arctan \left( { - \dfrac{1}{{\sqrt 3 }}} \right)$is $- \dfrac{\pi }{6}$.

Additional information:
1. In Mathematics the inverse trigonometric functions (every so often additionally called anti- trigonometric functions or cyclomatic function) are the reverse elements of the mathematical functions In particular, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are utilized to get a point from any of the point's mathematical proportions.
Reverse trigonometric functions are generally utilized in designing, route, material science, and calculation.

2. In inverse trigonometric function, the domain are the ranges of corresponding trigonometric functions and the range are the domain of the corresponding trigonometric function.
3. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
4. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.

Note:
1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2.Range of the inverse of tangent is in the interval $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and domain is $\left( { - \infty .\infty } \right)$.