In order to calculate the tangent of theta, let's first calculate the sine of theta using the following relation:
[tex]\begin{gathered} \sin ^2(\theta)+\cos ^2(\theta)=1 \\ \sin ^2(\theta)+(\frac{4}{9})^2=1 \\ \sin ^2(\theta)+\frac{16}{81}=1 \\ \sin ^2(\theta)=\frac{81}{81}-\frac{16}{81} \\ \sin ^2(\theta)=\frac{65}{81} \\ \sin (\theta)=\pm_{}\frac{\sqrt[]{65}}{9} \end{gathered}[/tex]Since pi < theta < 2pi, the sine is negative.
Now, calculating the tangent, we have:
[tex]\begin{gathered} \tan (\theta)=\frac{\sin (\theta)}{\cos (\theta)} \\ \tan (\theta)=\frac{-\frac{\sqrt[]{65}}{9}}{\frac{4}{9}} \\ \tan (\theta)=-\frac{\sqrt[]{65}}{4} \end{gathered}[/tex]