trig half angle identity question?

So the half angle formula for tan is:

tan(u/2)=(sin u) / (1+cos u)

tan 15 = tan (30/2)

And therefore:

tan (30/2) = (sin 30) / (1+cos 30)

tan (30/2) =(1/2) / [1 + (sqrt3) /2]

tan (30/2) =(1/2) / [2/2 + (sqrt3) /2]

tan (30/2) =1 / (2+sqrt3) = answer!

...(sqrt3) means square root of 3

If you write it out on paper it should make more sense visually.

that's oftentimes solved in 2 techniques one if x = one hundred fifty levels sin x = a million/2 cos x = -sqrt(3)/2 a million/2x = 75 levels tan x/2 = (a million-cos x)/sin x utilising the above = tan x/2 = 2(a million+sqrt(3)/2) = sqrt(3) +2 yet once you're no longer particlar a pair of million/2 attitude tan 75* = tan(40 5*+30*) = (tan 40 5+tan30)/(a million-tan 40 5 tan 30) tan 40 5 =a million tan 30 = a million/sqrt(3) so tan 75 = (a million+a million/sqrt(3))/(a million-sqrt(3) = (sqrt(3)+a million)/sqrt(3)(a million-sqrt(3) by utilising simplifying this you may get the consequence