Question

How do you use half angle formula to find `tan (5pi)/12`?

Answer 1

`tan(5pi) -= tan(pi) = 0`
and there is no point in using the half angle formula to evaluate
`(tan(5pi))/12`

Therefore, let's assume that you really want to evaluate
`tan((5pi)/12)`

Half angle formula (for tan):
`tan^2(theta/2) = (1-cos(theta))/(1+cos(theta))`

`tan^2((5pi)/12) = tan(((5pi)/6)/2)`

`=(1-cos(pi/6))/(1+cos(pi/6))`

`= (1+sqrt(3)/2)/(1-sqrt(3)/2)`

`= (1+sqrt(3)/2)^2/(1-3/4)`

`= 4(1+sqrt(3)/2)^2`

That is
`tan^2((5pi)/12) = (2(1+sqrt(3)/2))^2`

Since `(5pi)/12` is in Quadrant 1, the tan is positive
and
`tan((5pi)/12) = 2+sqrt(3)`