Question

Find the exact value: tan^-1(cot 2pi/3)

Answer 1

\[\cot \frac{ 2 \pi }{ 3 }=\cot \left( \frac{ \pi }{ 2 }+\frac{ \pi }{ 6 } \right)=-\tan \frac{ \pi }{ 6 }=\tan \left( \frac{ -\pi }{ 6 } \right)\]
\[\tan^{-1} \left( \cot \frac{ 2 \pi }{ 3 } \right)=\tan^{-1}( \tan( \frac{ -\pi }{ 6 }))=-\frac{ \pi }{ 6 }\]

Answer 2

Another option is just use a unit circle to look up the values

http://en.wikipedia.org/wiki/Unit_circle#mediaviewer/File:Unit_circle_angles_color.svg

\[\cot \frac{2\pi}{3} = \frac{\cos 2\pi/3}{\sin 2\pi/3} = -\frac{1}{\sqrt{3}}\]
\[\tan \theta = -\frac{1}{\sqrt{3}}\]
\[\theta = \frac{5\pi}{6} + \pi n\]

Answer 3

Thank you both very much!

Answer 4

yw