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
1/√3 = 2x/(1-x^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° = -√3-2
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