Given \(\cos x = -\frac{5}{13}\) where \(180^\circ < x < 270^\circ\): Using: \(\sin^2 x + \cos^2 x = 1\) \(\sin^2 x = 1 - \left(-\frac{5}{13}\right)^2 = \frac{144}{169} \implies \sin x = -\frac{12}{13}\) \(\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{5}\) Calculating \(\tan x - \sin x\): \(\tan x - \sin x = \frac{12}{5} - \left(-\frac{12}{13}\right) = \frac{12}{5} + \frac{12}{13}\) Finding a common denominator (65): \(\tan x - \sin x = \frac{12 \times 13}{65} + \frac{12 \times 5}{65} = \frac{156}{65} + \frac{60}{65} = \frac{216}{65}\) Thus, the value is: \(\tan x - \sin x = \frac{216}{65}\)