How do you find the exact value of ${tan}^{- 1} (- \frac{1}{\sqrt{3}})$ $tan^-1 (-1/sqrt3)$ ?

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Answer 1 · 2018-03-19T04:29:14

$θ = π n - (\frac{π}{6})$ $color(indigo)(theta = pi n - (pi/6)$ with $color(purple)(n in ZZ$

Let $theta = tan ^ -1 (-1/sqrt3)$

$tan theta = tan ( tan^ -1 (-1/ sqrt3)) = -(1/sqrt3)$

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From the above tables,
$tan (pi - (pi/6)) = tan -(pi/6) = -(1/ sqrt3)$

Generalising the same,

$color(indigo)(theta = pi n - (pi/6)$ with $color(purple)(n in ZZ$

How do you find the exact value of tan^-1 (-1/sqrt3)tan−1(−1√3)tan^-1 (-1/sqrt3)?