Given that $\cos \theta=-\frac{11}{61}$ and that angle $\theta$ terminates in quadrant III, then what is the value of $\tan \theta$ ?

Step 1 ：We are given that \(\cos \theta = -\frac{11}{61}\) and that angle \(\theta\) terminates in quadrant III.

Step 2 ：We know that the cosine of an angle is negative in the second and third quadrants. Since \(\theta\) is in the third quadrant, we can use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find the sine of \(\theta\).

Step 3 ：We know that the sine of an angle is positive in the first and second quadrants and negative in the third and fourth quadrants. Therefore, the sine of \(\theta\) will be negative.

Step 4 ：Once we have the sine of \(\theta\), we can find the tangent of \(\theta\) by using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Step 5 ：By substituting the values of \(\sin \theta\) and \(\cos \theta\) into the equation, we find that \(\tan \theta = 5.454545454545454\).

Step 6 ：Final Answer: The value of \(\tan \theta\) is \(\boxed{5.454545454545454}\).