trig
posted by Karen .
use a halfangle identity to find the exact value of tan 105 degrees.

we know that
tan 2A = 2tanA/(1  tan^2 A)
so tan 210 = 2tan105/(1  tan^2 105)
you should know that
tan 210 = tan 30° = 1/√3
let tan 105 = x
then
1/√3 = 2x/(1x^2)
2√3x = 1  x^2
x^2 + 2√3  1 = 0
x = (2√3 ± √16)/2
= √3 ± 2
but obviously tan105 is negative, ( 105° is in II )
so tan 105° = √32