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
