#tan ((11pi)/8) = tan ((3pi)/8 + (8pi)/8) = tan ((3pi)/8 + pi) = tan ((3pi)/8)#
Find #tan ((3pi)/8) = tan t#--> #tan 2t = tan ((3pi)/4) = - 1#
Use trig identity;
#tan 2t = (2tan t)/(1 - tan^2 t)#
#-1 = (2tan t)/(1 - tan^2 t)#
Cross multiply:
#tan^2 t - 1 = 2tan t#
#tan^2 t - 2tan t - 1 = 0#
Solve this quadratic equation for tan t:
#D = d^2 = b^2 - 4ac = 4 + 4 = 8# --> #d = +- 2sqrt2#
There are 2 real roots:
#tan t = -b/(2a) +- d/(2a) = 1 +- sqrt2#
#tan t = 1 + sqrt2 # and #tan t = 1 - sqrt2 #
Since #tan t = tan ((3pi)/8)# is positive, then
#tan ((11pi)/8) = tan ((3pi)/8) = 1 + sqrt2#
