using principal value find tan^-1[cot (pi/3)]

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Solution

$We know that the principal value branch of \tan^{- 1} is (\frac{- π}{2}, \frac{π}{2}) So \tan^{- 1} (\cot \frac{π}{3}), = \tan^{- 1} (\frac{1}{\sqrt{3}}), as \cot \frac{π}{3} = \frac{1}{\sqrt{3}} = \tan^{- 1} (\tan \frac{π}{6}) as \tan \frac{π}{6} = \frac{1}{\sqrt{3}} = \frac{π}{6} And this value lies within principal value branch of \tan^{- 1} so \tan^{- 1} (\cot \frac{π}{3}) = \frac{π}{6} .$

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