# How do you calculate ##Arctan( – sqrt 3/3)##?

##arctan(-sqrt(3)/3) = tan^-1(-sqrt(3)/3) = tanx = -sqrt(3)/3##

So basically what the problem asking you is that what angle on unit circle will give us ##-sqrt(3)/3## for tangent function.

You need to know the unit circle in order to find this the easy way, ##tan((pi)/6)## will give you ##sqrt(3)/3##. Finally put the negative for the above value.

## tan^-1(-sqrt(3)/3) = -pi/6##

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