First, find:
#tan ((113pi)/12) = 1/(cot t) = tan ((5pi)/12 + 9pi) =#
#= tan ((5pi)/12 + pi) = tan ((5pi)/12)#
Call # tan t = tan ((5pi)/12)# --> #tan 2t = tan ((10pi)/12) =
= tan ((5pi)/6) = - sqrt3/3# (trig table)
Use trig identity:
#tan 2t = - 1/sqrt3 = (2tan t)/(1 - tan^2 t) #
Cross multiply, then, solve the quadratic equation. for tan t
#tan^2 t - 1 = 2sqrt3tan t#
#tan^3 t - 2sqrt3tan t - 1 = 0#
#D = d^2 = b^2 - 4ac = 12 + 4 = 16# --> #d = +- 4#
There are 2 real roots:
#tan t = (2sqrt3)/2 +- 4/2 = sqrt3 +- 2.#
Since #t = (5pi)/12# is in Quadrant 1, then, tan t is positive.
Finally,
#cot ((113pi)/12) = 1/(tan t) = 1/(sqrt3 + 2)#
Check by calculator.
#t = (5pi)/12 = 75^@# --> #tan 75 = 3.732#
#sqrt3 + 2 = 1.732 + 2 = 3.732# Proved.
