Question

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

Answer 1

Hint: The range of ${\tan ^{ - 1}}\theta$ is between $\left. {\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right.} \right)$. From the trigonometric table find the value of$\dfrac{1}{{\sqrt 3 }}$ . Now substitute this back into our given expression and simplify it to get the principal value.

Complete step-by-step answer:
A principal value of a function is the value selected at a point in the domain of a multiple-valued function, chosen so that the function has a single value at the point. The principal value of ${\tan ^{ - 1}}\theta$
The principal value of ${\tan ^{ - 1}}\theta$ branches to,
${\tan ^{ - 1}}x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.
Hence the principal value of the given function will be between the range$\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ .
Now we have been given the function, ${\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)$.
Let us first find the value of $\dfrac{1}{{\sqrt 3 }}$from the above expression. By using the trigonometric table we can find the tangent function related to $\dfrac{1}{{\sqrt 3 }}$. Let us draw the trigonometric table to make the identification of the function easier.

From the table we get that, $\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}$.
Now let us substitute $\tan \dfrac{\pi }{6}$in the place of $\dfrac{1}{{\sqrt 3 }}$.Thus we can change the given expression as,

${\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) = {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{6}} \right)$
Now let us simplify the above expression to get the principal value.

$${\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) = {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{6}} \right) \ \\\ {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{6}} \right) = \dfrac{\pi }{6} \ \\\ {\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) = \dfrac{\pi }{6} \ \\\$$

Thus we got the principal value of the given inverse tangent function as, $\dfrac{\pi }{6}$.
$\therefore {\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) = \dfrac{\pi }{6}$.

Note: To solve a question like these you should be familiar with the domain and range of the sine functions as well as the domain and range of the inverse sine functions. The range of inverse tangent function is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and the domain of inverse function of tangent is $\left( { - \infty ,\infty } \right)$. Students should remember the important trigonometric ratios and standard angles to solve these types of questions.