#tan(195^o) = tan(390^o/2)#
The identity you need is:
#tan(x/2) = pmsqrt((1-cosx)/(1+cosx))#
#+# if quadrant I or III
#-# if quadrant II or IV
It's derived from:
#sin(x/2) = pmsqrt((1-cosx)/2)#
#+# if quadrant I or II
#-# if quadrant III or IV
#cos(x/2) = pmsqrt((1+cosx)/2)#
#+# if quadrant I or IV
#-# if quadrant II or III
Divide the quadrant conditions to get the sign you need. #tan390^o = tan30^o#, so you're in quadrant I. Thus, no matter what, the sign of the answer is positive.
#tan(390^o/2) = sqrt((1-cos390^o)/(1+cos390^o))#
#= sqrt((1-cos30^o)/(1+cos30^o))#
#= sqrt((1-sqrt3/2)/(1+sqrt3/2))#
Get common denominators:
#= sqrt(((2-sqrt3)/2)/((2+sqrt3)/2))#
Cancel:
#= sqrt((2-sqrt3)/(2+sqrt3))#
Multiply by the conjugate of the denominator:
#= sqrt(2-sqrt3)/sqrt(2+sqrt3)*sqrt(2-sqrt3)/sqrt(2-sqrt3)#
Simplify:
#= (2-sqrt3)/sqrt(2^2-(sqrt3)^2)#
#= (2-sqrt3)/sqrt(1)#
#= 2-sqrt3#
