How do you evaluate ${tan}^{- 1} (tan (\frac{11 π}{6}))$ $tan^-1(tan((11pi)/6))$ ?

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Answer 1 · 2016-08-31T04:53:57

11 $π$ $pi$ /6

say x = ${tan}^{- 1}$ $tan^-1$ tan(11 $π$ $pi$ /6)
so, tan x = tan(11 $π$ $pi$ /6)
or, x = 11 $π$ $pi$ /6

How do you evaluate tan−1(tan(11π6))tan^-1(tan((11pi)/6))?