tan(θ+60) = 2, Find tanθ.

*sqt3 = square root of 3

My attempt:

(tanΘ + tan60)/(1 - tanθtan60) = 2

tanΘ + (sqt3) = 2 - 2(sqt3)tanθ

tanΘ = (2 - (sqt3))/(2(sqt3) +1)

The answer is given as the reciprical of this which I think is wrong but could somebody check for me please?