Recall this identity

#sin^2(θ)+cos^2(θ)=1#

#sin^2(θ)=(-1/3)^2=1/9#

#1/9+cos^2(θ)=1#

#cos^2(θ)=9/9 - 1/9 = 8/9#

#cos(θ) = +-sqrt(8/9)=+- sqrt(8) / 3#

When we're between #180^@# and #270^@#, we're in the third quadrant, where sine and cosine are both negative. This means that we want the negative answer.

#cos(θ)=- sqrt(8) / 3#

So

#tan(θ)=sin(θ)/cos(θ)#

#tan(θ)=(-1/3) / (-sqrt(8) / 3)#

#tan(θ)=1/sqrt(8)" "# (negatives cancel out on division)

or

#tan(theta) = sqrt(8)/8 = sqrt(2)/4#