6 If \( \cos \theta=-\frac{5}{13} \) and \( \sin \theta>0 \), then \( \tan \theta \) is \( \begin{array}{llll}\text { 1) } \frac{5}{12} & \text { 2) }-\frac{5}{12} & \text { 3) } \frac{12}{5} & \text { 4) }-\frac{12}{5}\end{array} \)

To find \( \tan \theta \), we can use the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). We know \( \cos \theta = -\frac{5}{13} \). Since we are given that \( \sin \theta > 0 \), we are in the second quadrant where cosine is negative and sine is positive. First, we can use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \sin^2 \theta + \left( -\frac{5}{13} \right)^2 = 1 \] \[ \sin^2 \theta + \frac{25}{169} = 1 \] \[ \sin^2 \theta = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169} \] Taking the square root (and remembering that sine is positive in this quadrant), we find: \[ \sin \theta = \frac{12}{13} \] Now we can substitute \( \sin \theta \) and \( \cos \theta \) into the tangent formula: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{-5} = -\frac{12}{5} \] Thus, the answer is: \(-\frac{12}{5}\). The correct option is \( \text{4) } -\frac{12}{5} \).