How do you calculate ${tan}^{- 1} (tan (\frac{4 π}{3}))$ $tan^-1(tan((4pi)/3))$ ?

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Answer 1 · 2015-05-25T17:41:59

For any trig function, $T$ $T$

$T^{- 1} (T (θ)) = θ$ $T^-1(T(theta)) = theta$

Therefore ${tan}^{- 1} (tan (\frac{4 π}{3})) = \frac{4 π}{3}$ $tan^-1(tan((4pi)/3)) = (4pi)/3$

How do you calculate tan−1(tan(4π3))tan^-1(tan((4pi)/3))?