Question

Find the value of ${\tan }^{-1}\left(\tan \left(-6\right)\right).$

• A. $-6$
• B. $\pi -6$
• C. $-2\pi +6$
• D. $2\pi -6$

Answer 1

The correct option is D $2\pi -6$

$\mathrm{We}\mathrm{know}\mathrm{that}{\tan }^{-1}\left(\tan \theta \right)=\theta ,\mathrm{if}-\frac{\pi }{2}<\theta <\frac{\pi }{2}.\mathrm{Here},\theta =-6\mathrm{radians}\mathrm{which}\mathrm{does}\mathrm{not}\mathrm{lie}\mathrm{in}\mathrm{the}\mathrm{interval}\left(-\frac{\pi }{2},\frac{\pi }{2}\right).\mathrm{We}\mathrm{find}\mathrm{that}\left(2\pi -6\right)\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{interval}\left(-\frac{\pi }{2},\frac{\pi }{2}\right)\mathrm{such}\mathrm{that}\tan \left(2\pi -6\right)=-\tan 6=\tan \left(-6\right)$
$\therefore {\tan }^{-1}\left(\tan \left(-6\right)\right)={\tan }^{-1}\left(\tan \left(2\pi -6\right)\right\}=2\pi -6.$