Question

Question #75401

Answer 1

`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`