If sin theta = frac{-12}{13},cos phi = frac{- | vedclass

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(D) Given $\sin 	heta = \frac{-12}{13}$ and $	heta$ is in the third quadrant,$\cos 	heta = -\sqrt{1 - \sin^2 	heta} = -\sqrt{1 - (\frac{-12}{13})^

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$\frac{33}{56}$

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(D) Given $\sin 	heta = \frac{-12}{13}$ and $	heta$ is in the third quadrant,$\cos 	heta = -\sqrt{1 - \sin^2 	heta} = -\sqrt{1 - (\frac{-12}{13})^2} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{-12/13}{-5/13} = \frac{12}{5}$.Given $\cos \phi = \frac{-4}{5}$ and $\phi$ is in the third quadrant,$\sin \phi = -\sqrt{1 - \cos^2 \phi} = -\sqrt{1 - (\frac{-4}{5})^2} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.Thus,$	an \phi = \frac{\sin \phi}{\cos \phi} = \frac{-3/5}{-4/5} = \frac{3}{4}$.Using the formula $	an(	heta - \phi) = \frac{	an 	heta - 	an \phi}{1 + 	an 	heta 	an \phi}$:$	an(	heta - \phi) = \frac{12/5 - 3/4}{1 + (12/5)(3/4)} = \frac{(48 - 15)/20}{1 + 36/20} = \frac{33/20}{56/20} = \frac{33}{56}$.

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

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