$\sin \theta=-\frac{12}{13}, \tan \theta< 0$, find $\sec \theta$

Step 1 ：We are given that \(\sin \theta = -\frac{12}{13}\) and \(\tan \theta < 0\).

Step 2 ：Since the sine is negative and the tangent is negative, this means the angle is in the third quadrant. In the third quadrant, cosine is also negative.

Step 3 ：We can use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find the cosine of the angle.

Step 4 ：Substituting the given value of sine into the Pythagorean identity, we get \((-\frac{12}{13})^2 + \cos^2 \theta = 1\).

Step 5 ：Solving for \(\cos \theta\), we get \(\cos \theta = -\frac{5}{13}\).

Step 6 ：The secant of an angle is the reciprocal of the cosine of the angle. So, \(\sec \theta = -\frac{1}{\cos \theta}\).

Step 7 ：Substituting the value of \(\cos \theta\) into the equation, we get \(\sec \theta = -\frac{1}{-\frac{5}{13}}\).

Step 8 ：Solving for \(\sec \theta\), we get \(\sec \theta = -2.6\).

Step 9 ：So, the final answer is \(\boxed{-2.6}\).