#tan ((13pi)/12) = tan (pi/12 + pi) = tan pi/12 = (sin pi/12)/(cos pi/12)#
To find #sin (pi/12)# and #cos (pi/12)#, apply the 2 trig identities:
#cos 2a = 1 - 2sin^2 a#
#cos 2a = 2cos^2 a - 1#
#cos ((2pi)/12) = cos (pi/6) = sqrt3/2#
Call #sin (pi/12) = sin x#
a. #sqrt3/2 = 1 - 2sin^2 x#
#sin^2 x = 2 - sqrt3)/4#
#sin x = +- sqrt(2 - sqrt3)/2#.
Only the positive answer is accepted, since #pi/12# is in Quadrant I.
b.# sqrt3/2 = 2cos^2 x - 1#
#cos^2 x = (2 + sqrt3)/2#
#cos x = +- sqrt(2 + sqrt3)/2#. Only the positive answer is accepted.
#tan x = tan (pi/12) = sin/(cos) = sqrt(2 - sqrt3)/(sqrt(2 + sqrt3)#
