$\theta \in\left(3 \pi, \frac{7 \pi}{2}\right)$ i.e. 3 rd quadrant
$\begin{aligned} & \tan \theta=\frac{5}{12} \Rightarrow \sin \theta=\frac{-5}{13} \text { and } \cos \theta=\frac{-12}{13} \\ & \frac{\theta}{2} \in\left(\frac{3 \pi}{2}, \frac{7 \pi}{4}\right) \text { i.e. 4th quadrant } \\ & \cos \theta=2 \cos ^2 \frac{\theta}{2}-1 \Rightarrow \frac{-12}{13}=2 \cos ^2 \frac{\theta}{2}-1 \\ & \cos \frac{\theta}{2}=\frac{1}{\sqrt{26}} ; \therefore \sin \frac{\theta}{2}=-\frac{5}{\sqrt{26}} \\ & \Rightarrow \frac{\cos \frac{\theta}{2}-5 \sin \frac{\theta}{2}}{\sqrt{-(12 \sec \theta+5 \operatorname{cosec} \theta)}}=\frac{\frac{1}{\sqrt{26}}+\frac{25}{\sqrt{26}}}{\sqrt{-\left[\left(12 \cdot \frac{13}{-12}\right)+\left(5 \cdot \frac{13}{-5}\right)\right]}} \\ & =\frac{\sqrt{26}}{\sqrt{26}}=1 .\end{aligned}$