To find the tangent of the angle with a cosine of \(-\sqrt{17}\) in Quadrant III, we first need to determine the sine of the angle using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1. \]
Given that \(\cos \theta = -\sqrt{17}\):
\[ \sin^2 \theta + (-\sqrt{17})^2 = 1. \]
Calculating \((- \sqrt{17})^2\):
\[ \sin^2 \theta + 17 = 1. \]
Now, rearranging to find \(\sin^2 \theta\):
\[ \sin^2 \theta = 1 - 17 = -16. \]
This results in a negative value, which indicates a mistake in the setup since the cosine value should represent a valid real number. Since cosines generally lie between -1 and 1 for angles on the unit circle, let's clarify that \(\cos \theta = -\sqrt{17}\) is not a valid real cosine value.
Correcting this based on the usual context:
If we assume the given cosine was incorrectly noted, say we have:
For example, with \(cos(\theta) = -\frac{4}{5}\) or any value \(< 1\):
- Confirming with a valid \(\cos\) from generally real valued identities we could continue:
Continue solving the angle for sin value, then tangent from there.
Without any clear numerical value, solution cannot be derived using given cosine of \(-\sqrt{17}\).
Let's try reworking with an illustrative example close to valid:
Assuming cosine might instead be \(-\frac{3}{5}\):
- \(\sin^2 \theta + (-\frac{3}{5})^2 = 1 \implies sin^2 \theta + \frac{9}{25} = 1 \implies sin^2 \theta + \frac{9}{25} = \frac{25}{25} \implies sin^2 \theta = \frac{16}{25} \implies sin \theta = -\frac{4}{5}\)
Using this:
Then, \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}\).
If we calculate your original correctly, ensure the range and values are valid as analytical interpretation states.
For valid cosine/real computed angle, maintain fitness with note checking - clarify the angle setups!
Therefore, under reexamination the answer may not yield from negative square roots ascertainable through standard trigonometric identities. Review context of original prompt check on those values! ![So to conclude values can’t support ascertainable tangent, thus look confirmatively check back on conditions].