Answer
$-\frac{4}{3}$
Work Step by Step
$\sec^{2}\theta=1+\tan^{2}\theta$
$=1+(\frac{\sqrt 7}{3})^{2}=1+\frac{7}{9}=\frac{16}{9}$
$\implies \sec\theta=\pm\sqrt {\frac{16}{9}}=\pm\frac{4}{3}$
$\sec\theta$ is negative in the third quadrant.
$\implies \sec\theta=-\frac{4}{3}$
