Let $(-12,-5)$ be a point on the terminal side of $\theta$. Find the exact values of $\cos \theta, \csc \theta$, and $\tan \theta$. \[ \begin{array}{l} \cos \theta= \\ \csc \theta= \\ \tan \theta= \end{array} \]

Step 1 ：Given the point (-12,-5) on the terminal side of \(\theta\), we can determine that this point lies in the third quadrant of the Cartesian plane. In this quadrant, both cosine and sine are negative, and tangent is positive.

Step 2 ：The cosine of an angle in standard position is the x-coordinate of a point on the terminal side of the angle, the sine is the y-coordinate, and the tangent is the y-coordinate divided by the x-coordinate.

Step 3 ：Let's calculate the cosine, sine and tangent. For the x-coordinate, we have -12, and for the y-coordinate, we have -5.

Step 4 ：Using these coordinates, we can calculate \(\cos \theta = \frac{x}{\sqrt{x^2 + y^2}} = \frac{-12}{\sqrt{(-12)^2 + (-5)^2}} = -\frac{12}{13}\)

Step 5 ：Similarly, we can calculate \(\sin \theta = \frac{y}{\sqrt{x^2 + y^2}} = \frac{-5}{\sqrt{(-12)^2 + (-5)^2}} = -\frac{5}{13}\)

Step 6 ：Then, we can calculate \(\tan \theta = \frac{y}{x} = \frac{-5}{-12} = \frac{5}{12}\)

Step 7 ：Finally, we calculate \(\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{5}{13}} = -\frac{13}{5}\)

Step 8 ：So, the final answers are \(\cos \theta = \boxed{-\frac{12}{13}}\), \(\csc \theta = \boxed{-\frac{13}{5}}\), and \(\tan \theta = \boxed{\frac{5}{12}}\)