The value of ${\tan }^{-1}(\tan (-6\left)\right)$,is

Solution:

We know that ${\tan }^{-1}(\tan \theta )=\theta ,\text{ if }-\frac{\pi }{2}<\theta <\frac{\pi }{2}$

Here, $\theta =-6$ radians which does not lie between $-\frac{\pi }{2}\text{ and }\frac{\pi }{2}$

However $2\mathrm{\pi }-6$ lies between $-\frac{\pi }{2}\text{ and }\frac{\pi }{2}$ such that

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\tan (2\pi -6)=-\tan 6=\tan (-6)\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\tan }^{-1}\{\tan (-6\left)\right\}={\tan }^{-1}\{\tan (2\pi -6\left)\right\}=2\pi -6\end{array}$