tan^-1(-sqrt3/3)tan−1(−√33)
tan^-1xtan−1x means find the ANGLE that has a tangent of xx
The range of tan^-1tan−1 is -pi/2−π2 to pi/2π2
-sqrt3/3−√33 would fall in the fourth quadrant, so the value of tan^-1tan−1 is between -pi/2−π2 and 00 and is a negative angle.
Recall the identity tanx =sintheta/costhetatanx=sinθcosθ
Looking at the unit circle,
tan((11pi)/6)=frac{sin((11pi)/6)}{cos((11pi)/6)}=frac{-1/2}{sqrt3/2}=-1/2*2/sqrt3=-sqrt3/3tan(11π6)=sin(11π6)cos(11π6)=−12√32=−12⋅2√3=−√33
However, because the range of tan^-1tan−1 is pi/2π2 to -pi/2−π2,
the answer is -pi/6−π6 instead of (11pi)/611π6
