Question

Find the principal value of each of the following
${{\tan }^{-1}}\left( \cos \dfrac{\pi }{2} \right)$

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 positive 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= ${{\tan }^{-1}}\left( \cos \dfrac{\pi }{2} \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
$\tan y=\cos \dfrac{\pi }{2}=0$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Since zero is also considered as positive, 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 ${{\tan }^{-1}}\left( \cos \dfrac{\pi }{2} \right)$is
${{\tan }^{-1}}\left( \cos \dfrac{\pi }{2} \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Therefore,
$={{\tan }^{-1}}\left( \tan \left( 0 \right) \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
$=0$

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 $\dfrac{\pi }{2}$.