## Precalculus (6th Edition)

$\displaystyle \frac{\sqrt{5}}{5}$
$90^{o} < \theta < 180^{o}\quad/\div 2$ $45^{o} < \displaystyle \frac{\theta}{2} < 90^{o}$ so, $\theta$ is quadrant II where cos$\theta$ and sec$\theta$ are negative, $\displaystyle \frac{\theta}{2}$is in quadrant I, $\displaystyle \tan\frac{\theta}{2}$ and $\displaystyle \cot\frac{\theta}{2}$ are positive. Pythagorean identity: $\displaystyle \sec^{2}\theta=\tan^{2}\theta+1=\frac{5}{4}+1=\frac{9}{4}$ $\displaystyle \sec\theta=-\frac{3}{2}\Rightarrow\cos\theta=-\frac{2}{3}$ $\displaystyle \tan\frac{\theta}{2}=+\sqrt{\frac{1-(-\frac{2}{3})}{1+(-\frac{2}{3})}}=\sqrt{\frac{\frac{5}{3}}{\frac{1}{3}}}=\sqrt{5}$ So, cot and tan being reciprocal, $\displaystyle \cot\frac{\theta}{2}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$