Question

Find the principal value of ${{\tan }^{-1}}\left( -\sqrt{3} \right)$.

Answer 1

Hint: We will be using the concept of inverse trigonometric functions to solve the problem. We will first write $-\sqrt{3}$ in terms of tangent of an angle then we will use the fact that ${{\tan }^{-1}}\left( \tan x \right)=x$ for $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$.

Complete step-by-step answer:
Now, we have to find the value of${{\tan }^{-1}}\left( -\sqrt{3} \right)$.
Now, we will first represent it in terms of tangent of an angle. So, we know that the value of $-\sqrt{3}=\tan \left( -\dfrac{\pi }{3} \right)...........\left( 1 \right)$
We have taken $-\sqrt{3}=\tan \left( -\dfrac{\pi }{3} \right)$ as in the view of the principal value convention x is confined to$\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$.
Now, we know that the graph of ${{\tan }^{-1}}\left( \tan x \right)$ is,

Now, we have to find the value of${{\tan }^{-1}}\left( -\sqrt{3} \right)$.
We will use the equation (1) to substitute the value of $-\sqrt{3}$. So, we have,
${{\tan }^{-1}}\left( \tan \left( -\dfrac{\pi }{3} \right) \right)$
Also, we know that the value of ${{\tan }^{-1}}\left( \tan x \right)=x$. So, we have the value of,
${{\tan }^{-1}}\left( \tan \left( -\dfrac{\pi }{3} \right) \right)=-\dfrac{\pi }{3}$

Note: To solve these type of question it is important to note that we have used a fact that ${{\tan }^{-1}}\left( \tan x \right)=x$ only for $x$ belongs to $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. For another value of x the graph of ${{\tan }^{-1}}\left( \tan x \right)$ must be used to find the value.