How do you evaluate the equation $arctan (- \frac{\sqrt{3}}{3})$ $arctan( -sqrt3 / 3 )$ ?

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How do you evaluate the equation $arctan (- \frac{\sqrt{3}}{3})$ $arctan( -sqrt3 / 3 )$ ?

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Answer 1 · 2018-03-08T15:05:52

Converting the question in EQUATION FORM :
$x = arctan (- \frac{\sqrt{3}}{3}) = - \frac{π}{6}$ $color(red)(x)=arctan(-(sqrt3)/3)=-pi/6$

Explanation:

$(1) {tan}^{- 1} (- X) = - {tan}^{- 1} (X), X \in R$ $color(red)((1)tan^(-1)(-X)=-tan^-1(X),X in R$
$(2) tan (\frac{π}{6}) = \frac{1}{\sqrt{3}}$ $color(red)((2)tan(pi/6)=1/sqrt3$ .
$(3) {tan}^{- 1} (tan (X)) = X, \forall X \in (- \frac{π}{2}, \frac{π}{2})$ $color(red)((3)tan^(-1)(tan(X))=X,AA X in(-pi/2,pi/2))$
If the equation is, $x = {tan}^{- 1} (- \frac{\sqrt{3}}{3})$ $color(red)(x)=tan^(-1)(-(sqrt3)/3)$ , then applying (1) we get $x = - {tan}^{- 1} (\frac{\sqrt{3}}{3}) = - {tan}^{- 1} (\frac{1}{\sqrt{3}})$ $x=-tan^(-1)((sqrt3)/3)=-tan^(-1)(1/sqrt3)$ , now from (2) we get $x = - {tan}^{- 1} (tan (\frac{π}{6})), w h e r e, \frac{π}{6} \in (- \frac{π}{2}, \frac{π}{2})$ $x=-tan^(-1)(tan(pi/6)),where,pi/6 in(-pi/2,pi/2)$
$x = - \frac{π}{6} \to$ $x=-pi/6 to$ [ Applying (3) ]

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