How does one solve tan(75) by using a half -angle identity?

One half-angle identity for the tangent is given by

tan (a/2) = (1 - cos a) / sin a

So....a, in this case, is 150 degrees.   And we have

tan (75) = tan (150/2) = ( 1 - cos 150) / sin (150) = [1 - (-√3/2)] / (1/2) = [1 + √3/2] / (1/2) =

2 [ 2 + √3] / 2 = [ 2 + √3] ≈ 3.732

Note...the same result could have been obtained by using the tangent angle sum identity with the angles 30 degrees and 45 degrees.

I don't remember ever seeing this identity before!  

Thanks Chris.