Question 150901
Take note that {{{pi/6+5pi/3=11pi/6}}}



So 


{{{tan(11pi/6)=tan(pi/6+5pi/3)}}}




{{{tan(11pi/6)=(tan(pi/6)+tan(5pi/3))/(1-tan(pi/6)tan(5pi/3))}}} Use the Sum-Difference identity to rewrite the right side.



{{{tan(11pi/6)=((sqrt(3)/3)-sqrt(3))/(1-(sqrt(3)/3)(-sqrt(3)))}}} Using the unit circle, we get {{{tan(pi/6)=sqrt(3)/3}}} and {{{tan(5pi/3)=-sqrt(3)}}}



{{{tan(11pi/6)=((sqrt(3)/3)-sqrt(3))/(1-(-1))}}} Multiply {{{sqrt(3)/3}}} and {{{-sqrt(3)}}} to get {{{(sqrt(3)/3)(-sqrt(3))=-(sqrt(3)sqrt(3))/3=-3/3=-1}}}



{{{tan(11pi/6)=((sqrt(3)/3)-sqrt(3))/(2)}}} Combine like terms in the denominator.



{{{tan(11pi/6)=((sqrt(3)/3)-3sqrt(3)/3)/(2)}}} Multiply {{{-sqrt(3)}}} by {{{3/3}}} 



{{{tan(11pi/6)=(-2sqrt(3)/3)/(2)}}} Combine the fractions in the numerator.



{{{tan(11pi/6)=-sqrt(3)/3}}} Divide and simplify.




So the answer is {{{tan(11pi/6)=-sqrt(3)/3}}}