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

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

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Answer 1 · 2016-10-08T11:19:55

$- \frac{π}{6}$ $-pi/6$

Explanation:

Let $tan^-1(-sqrt3/3)= A :. tanA=(-sqrt3/3); tan(-pi/6)=-sqrt3/3 :.A= -pi/6$ .The range of $tan^-1x$ is $(-pi/2,pi/2)$
$:.tan^-1(-sqrt3/3)= -pi/6$ since $tan(-pi/6)=-sqrt3/3 and -pi/6 in (-pi/2,pi/2)$ [Ans]

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

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