Question

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

Answer 1

let us consider the y as given inverse trigonometric function and then we will get the value of $\tan y$ and if $\tan y$ is negative then the principal value will be $-\theta$. The principal value of $\tan \theta$ lies between $-\dfrac{\pi }{2}$and $\dfrac{\pi }{2}$

Complete step-by-step answer:
Let y= ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
$\tan y=-\dfrac{1}{\sqrt{3}}$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Since $-\dfrac{1}{\sqrt{3}}$is negative, principal value is $-\theta$
We know the principal value of ${{\tan }^{-1}}$is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$
Hence the principal value of ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$is
$={{\tan }^{-1}}\left( -\tan \dfrac{\pi }{6} \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
We know that $\tan \left( -\theta \right)=-\tan \theta$
Therefore,
$={{\tan }^{-1}}\left( \tan \left( -\dfrac{\pi }{6} \right) \right)$. . . . . . . . . . . . . . . . . . . . . (4)
$=-\dfrac{\pi }{6}$

Note: The general formula for principal value of ${{\tan }^{-1}}\left( \cot \theta \right)=\dfrac{\pi }{2}-\theta$ if and only if $\left( 0,\pi \right)$ . The principal value of $\theta$ for any given inverse function should lie within its range. The range of inverse tangent function or arc tangent function is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. So the principle of $\theta$ for inverse tangent function always lies between $-\dfrac{\pi }{2}$and $\dfrac{\pi }{2}$.