Use the fact that tangent is an odd function. This means
tan(-pi/12) = -tan(pi/12)tan(−π12)=−tan(π12)
Use the Half Angle Identity -tan(theta/2) = -sqrt((1-cos theta)/(1+cos theta))−tan(θ2)=−√1−cosθ1+cosθ:
Let theta = pi/6 => tan((pi/6)/2) = tan(pi/6 * 1/2) = tan(pi/12)θ=π6⇒tan(π62)=tan(π6⋅12)=tan(π12)
-tan((pi/6)/2) = -sqrt((1-cos( pi/6))/(1+cos (pi/6))−tan(π62)=−
⎷1−cos(π6)1+cos(π6)
cos (pi/6) = cos (30^@) = sqrt(3)/2cos(π6)=cos(30∘)=√32
-tan((pi/6)/2) = -sqrt((1- sqrt(3)/2)/(1 + sqrt(3)/2))−tan(π62)=−
⎷1−√321+√32
Find common denominators:
tan(-pi/12) = -sqrt(((2-sqrt(3))/2)/((2+sqrt(3))/2)) = - sqrt(((2-sqrt(3))/cancel(2)) * cancel(2)/(2+sqrt(3)))
tan(-pi/12) = -sqrt((2-sqrt(3))/(2+sqrt(3))
Multiply by the conjugate:
tan(-pi/12) = -sqrt((2-sqrt(3))/(2+sqrt(3)) * (2-sqrt(3))/(2-sqrt(3))
tan(-pi/12) = - sqrt((2-sqrt(3))^2/(4 - 3)) = - (2-sqrt(3))
tan(-pi/12) =sqrt(3) - 2
