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Find the value of $\; \tan^{-1}\bigg(\tan{\frac{5\pi}{6}}\bigg)+\cos^{-1}\bigg(\cos{\frac{13\pi}{6}}\bigg)$

$\begin{array}{1 1} 0 \\ 3 \pi \\ \pi \\ \frac{\pi}{3} \end{array} $

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  • Principal interval of tan is \((-\large\frac{\pi}{2},\large\frac{\pi}{2})\)
  • Principal interval of cos is \([0,\pi]\)
  • \(-tan\theta=tan(-\theta)\)
  • \(tan(\pi-\theta)=-tan\theta\)
  • \(cos(2\pi+\theta)=cos\theta\)
Since \(\large\frac{5\pi}{6}\)is not within the principal interval  reduce it within the principal interval of tan.
Similarly reduce \(\large\frac{13\pi}{6}\) within the principal interval of cos
By taking \(\theta=\large\frac{\pi}{6},\:tan(\pi-\large\frac{\pi}{6})=-\tan\large\frac{\pi}{6}=tan(-\large\frac{\pi}{6})\:and\)
\(\Rightarrow\:tan\large\frac{5\pi}{6}=tan(-\large\frac{\pi}{6})\) and
\( tan^{-1} \bigg[ tan \bigg( \pi - \large\frac{\pi}{6} \bigg)\bigg] + cos^{-1} \bigg( cos (2\pi+\large\frac{\pi}{6} )\bigg) \)
\( = \large\frac{-\pi}{6}+\large\frac{\pi}{6} = 0\)


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