Call tan ((-pi)/12) = tan x.
tan 2x = tan ((-2pi)/12) = - tan (pi/6) = - sqrt3/3 = - 1/sqrt3
Apply the trig identity: tan 2x = (2tan x)/(1 - tan^2 x), we get:
-1/sqrt3 = (2tan x)/(1 - tan^2 x). Cross multiply:
tan^2 x - 1 = 2sqrt3tan x.
Solve the quadratic equation:
tan^2 x - 2sqrt3tan x - 1 = 0
D = d^2 = b^2 - 4ac = 12 + 4 = 16 --> d = +- 4
tan x = (2sqrt3)/2 +- 4/2 = sqrt3 +- 2.
Since the arc (-pi/12) is in Quadrant IV, its tan is negative, then:
tan x = tan ((-pi)/12) = sqrt3 - 2.
Check by calculator.
tan ((-pi)/12) = tan (-15^@) = - 0.27.
(sqrt3 - 2) = (1.73 - 2) = - 0.27. OK
