Algebraically determine the exact value of?

We seek:

# tau= tan((11pi)/12) #

Using the trigonometric identity:

# tan(A+-B) -= (tanA+-tanB)/(1 \ bar("+") \ tanAtanB) # ..... [A]

We can write

# tau = tan(pi-pi/12) #

# \ \ = (tan(pi)-tan(pi/12))/(1+tan(pi)tan(pi/12)) #

# \ \ = (0-tan(pi/12))/(1+0) #

# \ \ = -tan(pi/12) #

And we can also use the same identity [A] to give:

# tan(2 (pi/12)) = (tan(pi/12)+tan(pi/12))/(1 - tan(pi/12)tan(pi/12)) #

# :. tan(pi/6) = (2tan(pi/12))/(1 - tan^2(pi/12) #

And we know that #tan(pi/6) = sqrt(3)/3# and #tan(pi/12) = -tau# so:

# sqrt(3)/3 = (-2tau)/(1 - (-tau)^2) #

# :. 1 - tau^2= -6/sqrt(3)tau #

# :. 1 - tau^2= -2sqrt(3)tau #

# :. tau^2 - 2sqrt(3)tau - 1 = 0#

Which is quadratic in #tau#, so we can complete the square:

# :. (tau - sqrt(3))^2 - (sqrt(3))^2 - 1 = 0#

# :. (tau - sqrt(3))^2 - 3 - 1 = 0#

# :. (tau - sqrt(3))^2 = 4 #

# :. tau - sqrt(3) = +-2 #

# :. tau = +-2 + sqrt(3)#

Noting that #(11pi)/12# is in #QII#, then we know that # tau= tan((11pi)/12) # is negative, thus we deduce that

# tau = -2 + sqrt(3)#

#tan ((11pi)/12) = tan (-pi/12 + pi) = tan (-pi/12) = - tan (pi/12)#
First, find #tan (pi/12)# by using the trig identity:
#tan 2t = (2tan t)/(1 - tan^2 t)#
Call #tan (pi/12) = tan t#, --> #tan 2t = tan (pi/6) = 1/sqrt3#
we get:
#(2tan t)/(1 - tan^2 t) = 1/sqrt3#
Cross multiply
#2sqrt3tan t = 1 - tan^2 t#
#tan^2 t + 2sqrt3tan t - 1 = 0#.
Solve this quadratic equation for tan t:
#D = d^2 = b^2 - 4ac = 12 + 4 = 16# --> #d = +- 4#
The 2 real roots are:
#tan t = -b/(2a) +- d/(2a) = - sqrt3 +- 2#
a. #tan t = tan (pi/12) = -2 - sqrt3# (rejected since #tan (pi/12) > 0#)
b. #tan t = tan (pi/12) = 2 - sqrt3#
Finally,
#tan ((11pi)/12) = - tan (pi/12) = sqrt3 - 2#