Question

How do you find the exact value of tan 5pi/12 ?

I've worked through it many times but keep ending up with this,
`(sqrt3/3+1)/(1-sqrt3/3)`
is this correct? and how do I simplify it?

Answer 1

(2 + sqrt3)

Explanation:

Use trig table of special arcs, unit circle, property of complement arcs:
`tan ((5pi)/12) = tan ((6pi)/12 - pi/12) = tan (pi/2 - (pi)/12) = cot (pi/12) = 1/(tan (pi/12)` (1)
First, find `tan (pi/12)`. Call `tan (pi/12) = tan t` --->
`tan 2t = tan (pi/6) = 1/sqrt3`
Use trig identity: `tan 2t = (2tan t)/(1 - tan^2 t)`.
In this case:
`(2tan t)/(1 - tan^2 t) = 1/sqrt3`
`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`
There are 2 real roots:
`tan t = - sqrt3 +- 2`.
Since `tan (pi/12)` is positive, take the positive value.
`tan t = tan (pi/12) = 2 - sqrt3`.
Back to equation (1) -->
`tan ((5pi)/12) = 1/(tan (pi/12)) = 1/(2 -sqrt3) =`
Multiply both numerator and denominator by `(2 - sqrt3)`
`tan ((5pi)/12) = (2 + sqrt3)/(4 - 3) = 2 + sqrt3`