Question

How do you evaluate the value of $\arctan \left( \sqrt{3} \right)$?

Answer 1

We start solving the problem by recalling the fact that $\arctan \left( x \right)={{\tan }^{-1}}\left( x \right)$. We then assume the $\arctan \left( \sqrt{3} \right)$ equal to a variable and then apply tangent on the both sides of it. We then make use of the fact that $\tan \left( {{\tan }^{-1}}a \right)=a$ to proceed through the problem. We then 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 $\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ to get the required answer.

Complete step-by-step solution:
According to the problem, we are asked to find the value of $\arctan \left( \sqrt{3} \right)$.
We know that $\arctan \left( x \right)={{\tan }^{-1}}\left( x \right)$. So, we get $\arctan \left( \sqrt{3} \right)={{\tan }^{-1}}\left( \sqrt{3} \right)$.
Let us assume ${{\tan }^{-1}}\left( \sqrt{3} \right)=\alpha$ ---(1).
Let us apply tangent on both sides of the equation (1).
$\Rightarrow \tan \left( {{\tan }^{-1}}\left( \sqrt{3} \right) \right)=\tan \alpha$ ---(2).
We know that $\tan \left( {{\tan }^{-1}}a \right)=a$, for $a\in \mathbb{R}$. Let us use this result in equation (2).
$\Rightarrow \sqrt{3}=\tan \alpha$ ---(3).
We know that if $\tan \theta =a$, for $a\in \mathbb{R}$, then the value of the angle $\theta$ lies in the interval $\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$.
So, we have $\tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}$, as the angle must lie in the interval $\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$. Let us use this result in equation (3).
$\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\tan \alpha$ ---(4).
We know that if $\tan \theta =\tan \alpha$, where $\theta \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$, then the principal solution of the angle $\alpha$ is equal to $\theta$. Now, let us use this result in equation (4).
$\Rightarrow \alpha =\dfrac{\pi }{3}$.
$\therefore$ We have found the value of $\arctan \left( \sqrt{3} \right)$ as $\dfrac{\pi }{3}$.

Note: Here we have assumed that we need to find the principal solution for the value $\arctan \left( \sqrt{3} \right)$. Whenever we get this type of problem, we first check whether the solution needs to be the principal solution or the general solution. We should only report the angle that was present in the principal range of the inverse of the tangent function. Similarly, we can expect problems to find the general solution of the value $\arctan \left( \dfrac{1}{\sqrt{3}} \right)$.