Question #75401

1 Answer
Nghi N
May 19, 2017

#sqrt3 + 2#

Explanation:

Use trig identity:
#tan 2a = (2tan a)/(1 - tan^2 a)#
Call tan (pi/12) = tan t
tan 2t = tan (pi/6) = 1/sqrt3
tan 2t = 1/sqrt3 = (2tan t)/(1 - tan^2 t)
Cross multiply:
#1 - tan^2 t = 2sqrt3tan t#.
Solve quadratic equation for tan t:
#- tan^2 t + 2sqrt3tan t + 1 = 0#
#D = d^2 = b^2 - 4ac = 12 + 4 = 16# --> #d = +- 4#
There are 2 real roots:
#tan (pi/12) = tan t = -b/(2a) +- d/(2a) = sqrt3 +- 2#
Since tan (pi/12) is positive, take the positive answer:
#tan (pi/12) = sqrt3 + 2#

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