#### Answer

$-\sqrt{7}$

#### Work Step by Step

$180^{o} < \theta < 270^{o}\quad/\div 2$
$90^{o} < \displaystyle \frac{\theta}{2} < 135^{o}$,
so, $\theta$ is quadrant III where cos (sec) and sin are negative,
$\displaystyle \frac{\theta}{2} $is in quadrant II, $\displaystyle \tan\frac{\theta}{2}$ is negative.
In the Half-Angle Identity
$\displaystyle \tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$,
... the sign is negative, we need $\displaystyle \cos\theta=\frac{1}{\sec\theta}$
Pythagorean identity:
$\displaystyle \sec^{2}\theta=\tan^{2}\theta+1=\frac{7}{9}+1=\frac{16}{9}$
$\displaystyle \sec\theta=-\frac{4}{3}\Rightarrow\cos\theta=-\frac{3}{4}=-0.75$
So,
$\displaystyle \tan\frac{\theta}{2}=-\sqrt{\frac{1-(-0.75)}{1+(-0.75)}}=-\sqrt{\frac{1.75}{0.25}}$
$=-\sqrt{7}$