\[ \cos \theta = \frac{5}{13}, \quad \frac{3\pi}{2} < \theta < 2\pi \] From the above, \(\theta\) lies in the fourth quadrant. So, \(\tan \theta\) will be negative. \[ \text{Hypotenuse} = 13 \] \[ \text{Base} = 5 \] \[ \text{Perpendicular} = \sqrt{13^2 + 5^2} = \pm 12 \] \[ \tan \theta = \frac{\text{perpendicular}}{\text{base}} \] \[ \tan \theta = \frac{-12}{5}, \quad \textbf{tan } \theta \textbf{ is negative in the fourth quadrant} \] \[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \] \[ \tan 2\theta = \frac{2 \times \frac{-12}{5}}{1 - \frac{(-12)^2}{5^2}} \] \[ \tan 2\theta = \frac{120}{119} \]