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Find the value of the following: \[ tan^{-1} \bigg(tan \frac{7\pi}{6}\bigg)\]

$\begin{array}{1 1} \frac{\pi}{6} \\ \frac{7 \pi}{6} \\ \frac{-7 \pi}{6} \\ \frac{-\pi}{6} \end{array} $

1 Answer

  • Principal interval of tan is\((\large-\frac{\pi}{2},\large\frac{\pi}{2})\)
  • \(tan(\pi+x)=tanx\)
Given $tan^{-1} \bigg(tan \large\frac{7\pi}{6}\bigg)$
\(\large\frac{7\pi}{6}\) is not in the principal interval of $tan \rightarrow$ reduce it within the principal interval $\rightarrow$ \(\large\frac{7\pi}{6}=\pi+\large\frac{\pi}{6}\)
The given expression can be written as \( tan^{-1}tan \bigg( \pi+\large\frac{\pi}{6} \bigg) \)
By taking \( x=\large\frac{\pi}{6},\) tan\((\pi+\large\frac{\pi}{6})=tan\large\frac{\pi}{6}\)
\( \Rightarrow tan^{-1}tan \large\frac{\pi}{6}=\large\frac{\pi}{6}\)


answered Feb 23, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1