If \(\sin \theta=\frac{-12}{13}, \cos \phi=\frac{-4}{5}\) and \(\theta, \phi\) lie in the third quadrant, then \(\tan(\theta-\phi)=\)

(D)
\(\sin \theta=\frac{-12}{13} \Rightarrow \cos \theta=\sqrt{1-\left(\frac{-12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\pm \frac{5}{13}\)
\(\because \theta\) is in third quadrant \(\Rightarrow \cos \theta \quad=-\frac{5}{13} \Rightarrow \frac{\sin \theta}{\cos \theta}=\tan \theta=\frac{12}{5}\)
\(\cos \phi=-\frac{4}{5} \Rightarrow \sin \phi=\sqrt{1-\left(\frac{-4}{5}\right)^{2}}=\sqrt{\frac{9}{25}}=\pm \frac{3}{5}\)
\(\phi\) is in third quadrant \(\Rightarrow \sin \phi=\frac{-3}{5} \Rightarrow \frac{\sin \phi}{\cos \phi}=\tan \phi=\frac{3}{4}\) \(\tan (\theta-\phi)=\frac{\tan \theta-\tan \phi}{1+\tan \phi \cdot \tan \phi}\)
\(=\frac{\frac{12}{5}-\frac{3}{4}}{1+\left(\frac{12}{5} \times \frac{3}{4}\right)}=\frac{\frac{33}{20}}{\frac{56}{20}}=\frac{33}{56}\)