Taking the squared norm of the numerator and denominator separately, $ \eqalign{ \left|e^{ 2i\theta}\pm e^{ i\theta}-1\right|^2 &= \left(e^{ 2i\theta}\pm e^{ i\theta}-1\right)\cdot \left(e^{-2i\theta}\pm e^{-i\theta}-1\right)\\ & \matrix{=& 1 & \pm2e^{ i\theta} & -e^{2i\theta} \\\\ & \pm2e^{-i\theta} & +4 & \mp2e^{ i\theta} \\\\ & -e^{2i\theta} & \mp2e^{-i\theta} & +1 } \\\\ &= 6 - 2\cos 2\theta \pm 2\cos\theta \mp 2\cos\theta \\\\ &= 6 - 2\cos 2\theta\,. } $ Notice, however, that this no longer depends on the sign, i.e. it is the same for the numerator and denominator.
But I admit, I like @Raymond's and @Aryabhata's answers much better!