If 180° < θ < 270°, then 90° < θ/2 < 135°, which places θ/2 in the second quadrant so that sin(θ/2) > 0 and cos(θ/2) < 0.
Recall that
cos²(θ/2) = (1 + cos(θ))/2
==> cos(θ/2) = -√[(1 + (-15/17))/2] = -1/√17
and
sin²(θ/2) = (1 - cos(θ))/2
==> sin(θ/2) = +√[(1 - (-15/17))/2] = 4/√17
Then
tan(θ/2) = sin(θ/2) / cos(θ/2)
… = (4/√17) / (-1/√17)
… = -4
