Recall that
[tex]\sin^2\theta+\cos^2\theta=1[/tex]
So we have
[tex]\sin^2\theta=1-\cos^2\theta=1-\left(\dfrac{\sqrt2}2\right)^2=1-\dfrac24=\dfrac12[/tex]
So either [tex]\sin\theta=\pm\dfrac1{\sqrt2}=\pm\dfrac{\sqrt2}2[/tex].
Since [tex]\dfrac{3\pi}2<\theta<2\pi[/tex], we expect [tex]\sin\theta<0[/tex], so we take the negative square root. Then
[tex]\sin\theta=-\dfrac{\sqrt2}2[/tex]
This then means we have
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt2}2}{\frac{\sqrt2}2}=-1[/tex]
