A derivation for #tan(x/2)# is at the bottom.
#tan(x/2) = pmsqrt((1-cosx)/(1+cosx))#
depending on the quadrant of #105^o# (quadrant II, thus the horizontal axis #< 0# on the unit circle).
#tan 105^o = tan ((7pi)/12) = tan (1/2*(7pi)/6) = -sqrt((1-cos((7pi)/6))/(1+cos((7pi)/6)))#
#= -sqrt((1+sqrt3/2)/(1-sqrt3/2))#
#= -sqrt(((2+sqrt3)/2)/((2-sqrt3)/2))#
#= -sqrt((2+sqrt3)/(2-sqrt3))#
#= -sqrt((2+sqrt3)^2/((2-sqrt3)(2+sqrt3)))#
#= -sqrt((2+sqrt3)^2)/(1)#
And since we already specified the quadrant, there's no need for #pm# (and #2 + sqrt3 > 0# of course).
#= -(2 + sqrt3)#
#= -2 - sqrt3#
You can derive the half-angle formula if you don't remember it.
#sin^2(x) = (1-cos(2x))/2#
Similarly:
#sin^2(x/2) = (1-cos(x))/2#
Thus:
#|sin(x/2)| = sqrt((1-cosx)/2)#
Similarly:
#cos^2(x) = (1+cos(2x))/2#
#cos^2(x/2) = (1+cosx)/2#
#|cos(x/2)| = sqrt((1+cosx)/2)#
Now we can divide them.
#|sin(x/2)/(cos(x/2))| = |tan(x/2)| = sqrt((1-cosx)/2) / sqrt((1+cosx)/2)#
#= sqrt((1-cosx)/(1+cosx))#
