Answer
$\dfrac{2\sqrt{30}-1}{\sqrt{15}+2\sqrt 2}$
Work Step by Step
Here, we have $\tan(\theta+\phi)=\dfrac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}$
Since, $\cos \phi$ is negative in third quadrant. so, we have $\cos \theta=\dfrac{-1}{3}$ and $\sin \theta=-\dfrac{2\sqrt 2}{3}$
$\tan(\theta+\phi)=\dfrac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}$
Then, we get
$\dfrac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\dfrac{2\sqrt 2-\dfrac{1}{\sqrt {15}}}{1-\dfrac{2 \sqrt 2}{\sqrt {15}}}$
or, $=\dfrac{2\sqrt{30}-1}{\sqrt{15}+2\sqrt 2}$
