## Precalculus (6th Edition)

$-\sqrt{7}$
$180^{o} < \theta < 270^{o}\quad/\div 2$ $90^{o} < \displaystyle \frac{\theta}{2} < 135^{o}$, 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, $\displaystyle \tan\frac{\theta}{2}=-\sqrt{\frac{1-(-0.75)}{1+(-0.75)}}=-\sqrt{\frac{1.75}{0.25}}$ $=-\sqrt{7}$