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

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

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Answer 1 · 2015-04-25T12:18:31

Assuming by ${tan}^{- 1} (\sqrt{3})$ $tan^(-1)(sqrt(3))$ you mean $arctan (\sqrt{3})$ $arctan(sqrt(3))$

and not $\frac{1}{tan (\sqrt{3})}$ $1/(tan(sqrt(3))$

We are looking for a value $θ$ $theta$
such that
$tan (θ) = \sqrt{3}$ $tan(theta) = sqrt(3)$
that is
$θ = arctan (tan (θ)) = arctan (\sqrt{3})$ $theta = arctan(tan(theta)) = arctan(sqrt(3))$

$tan (θ) = \sqrt{3}$ $tan(theta) = sqrt(3)$
is the $tan$ $tan$ of a standard triangle
with $θ = \frac{π}{3}$ $theta = pi/3$

That is $arctan (\sqrt{3}) = P \frac{I}{3}$ $arctan(sqrt(3)) = PI/3$
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How do you calculate ${tan}^{- 1} (\sqrt{3})$ $tan^-1(sqrt3)$ ?

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