Question

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

Answer 1

Hint: 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 \[\dfrac{\pi }{2}\].