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Find the value of $\tan^{-1}\bigg(\tan{\frac{2\pi}{3}}\bigg)$

$\begin{array}{1 1} \frac{ 4 \pi}{3} \\ \frac{2\pi}{3} \\ \frac{- \pi}{3} \\ \frac{\pi}{3} \end{array} $

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  • Principal interval of tan is \((-\large\frac{\pi}{2},\large\frac{\pi}{2})\)
  • \(tan(\pi-\theta)=-tan\theta\)
  • \(-tan\theta=tan(-\theta)\)
\(\large\frac{2\pi}{3}\) is not in the principal interval of tan
Reduce is and bring it within the principal interval
By taking \(\theta=\large\frac{\pi}{3}\) in tan\((\pi-\theta)\) we get
\( tan^{-1}tan\large\frac{2\pi}{3} \)
\( = tan^{-1} \bigg( tan \bigg( \pi -\large \frac{\pi}{3} \bigg) \bigg) \)
\( = tan^{-1} tan(- \frac{\pi}{3} ) = \large\frac{-\pi}{3} \)


answered Feb 19, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
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