How do you evaluate ${tan}^{- 1} (\frac{\sqrt{3}}{3})$ $tan^-1(sqrt3/3)$ ?

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Answer 1 · 2016-08-27T17:31:12

${tan}^{- 1} (\frac{\sqrt{3}}{3}) = \frac{π}{6}$ $tan^-1(sqrt3/3)=pi/6$ .

The Defn. of ${tan}^{- 1}$ $tan^-1$ function is :

$tan^-1x=theta, x in RR iff tantheta=x, theta in (-pi/2, pi/2)$

As $tan (pi/6)=1/sqrt3=sqrt3/3, and, pi/6 in (-pi/2, pi/2)$ , we get,

$tan^-1(sqrt3/3)=pi/6$ .

How do you evaluate tan^-1(sqrt3/3)tan−1(√33)tan^-1(sqrt3/3)?