Find $\sin \theta$. \[ \sec \theta=\frac{13}{6}, \tan \theta<0 \] \[ \sin \theta= \] (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Step 1 ：We know that \(\sec \theta = \frac{1}{\cos \theta}\), so \(\cos \theta = \frac{1}{\sec \theta} = \frac{6}{13}\).

Step 2 ：We also know that \(\sin^2 \theta + \cos^2 \theta = 1\), so we can solve for \(\sin \theta\) by rearranging this equation to get \(\sin \theta = \sqrt{1 - \cos^2 \theta}\).

Step 3 ：However, we also know that \(\tan \theta < 0\), which means that \(\sin \theta\) and \(\cos \theta\) must have opposite signs. Since \(\cos \theta > 0\), \(\sin \theta\) must be negative.

Step 4 ：Therefore, \(\sin \theta = -\sqrt{1 - \cos^2 \theta}\).

Step 5 ：Final Answer: \(\sin \theta = \boxed{-\frac{115}{130}}\)