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))
