How do you evaluate ${tan}^{- 1} (\frac{\sqrt{3}}{3})$ $tan^-1(sqrt3/3)$ without a calculator?

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Answer 1 · 2016-09-29T21:23:21

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

Consider a triangle with sides in ratio $1 : \sqrt{3} : 2$ $1:sqrt(3):2$

This is a right angled triangle since:

$1^{2} + {\sqrt{3}}^{2} = 1 + 3 = 4 = 2^{2}$ $1^2+sqrt(3)^2 = 1+3 = 4 = 2^2$

In fact, it is one half of an equilateral triangle, hence has angles $\frac{π}{6}$ $pi/6$ , $\frac{π}{3}$ $pi/3$ , $\frac{π}{2}$ $pi/2$ ...

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Now $tan (θ) = \frac{opposite}{adjacent}$ $tan(theta) = "opposite"/"adjacent"$

Hence we find:

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

Since $\frac{π}{6}$ $pi/6$ radians is in Q1, this is the value of ${tan}^{- 1} (\frac{\sqrt{3}}{3})$ $tan^(-1) (sqrt(3)/3)$

How do you evaluate tan−1(√33)tan^-1(sqrt3/3) without a calculator?