Answer
$\displaystyle \frac{\sqrt{5}}{5}$
Work Step by Step
$90^{o} < \theta < 180^{o}\quad/\div 2$
$45^{o} < \displaystyle \frac{\theta}{2} < 90^{o}$
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,
$\displaystyle \cot\frac{\theta}{2}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$
