To determine the value of [tex]\(\cos (\theta)\)[/tex] given the equation [tex]\(\tan (\theta) = -\sqrt{\frac{19}{17}}\)[/tex] and knowing that [tex]\(\theta\)[/tex] is an angle in quadrant II, let's follow these steps:
1. Understand the relationship between sine, cosine, and tangent:
[tex]\[
\tan (\theta) = \frac{\sin (\theta)}{\cos (\theta)}
\][/tex]
2. Understand that in quadrant II:
- The sine function ([tex]\(\sin (\theta)\)[/tex]) is positive.
- The cosine function ([tex]\(\cos (\theta)\)[/tex]) is negative.
3. Use the given information [tex]\(\tan (\theta) = -\sqrt{\frac{19}{17}}\)[/tex]:
- This means that [tex]\(\frac{\sin (\theta)}{\cos (\theta)} = -\sqrt{\frac{19}{17}}\)[/tex].
4. Identify the Pythagorean identity:
- [tex]\(\sin^2 (\theta) + \cos^2 (\theta) = 1\)[/tex].
5. Express [tex]\(\sin (\theta)\)[/tex] and [tex]\(\cos (\theta)\)[/tex] in terms of the given tangent value:
[tex]\[
\left( \frac{\sin (\theta)}{\cos (\theta)} \right)^2 = \left( -\sqrt{\frac{19}{17}} \right)^2 = \frac{19}{17}
\][/tex]
6. Assume [tex]\(\sin(\theta) = \sqrt{A}\)[/tex] and [tex]\(\cos(\theta) = -\sqrt{B}\)[/tex], and then:
[tex]\[
\left( \frac{\sqrt{A}}{-\sqrt{B}} \right)^2 = \frac{A}{B} = \frac{19}{17}.
\][/tex]
This implies:
[tex]\[
A = 19 \quad \text{and} \quad B = 17.
\][/tex]
7. Use the Pythagorean identity for sine and cosine:
[tex]\[
\sin^2 (\theta) + \cos^2 (\theta) = 1.
\][/tex]
Substituting [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[
\frac{19}{19 + 17} + \frac{17}{19 + 17} = 1.
\][/tex]
8. Determine [tex]\(\cos (\theta)\)[/tex]:
[tex]\[
\cos (\theta) = -\sqrt{\frac{17}{36}} = -\frac{\sqrt{17}}{6}.
\][/tex]
The correct answer is:
[tex]\[
\boxed{-\frac{\sqrt{17}}{6}}
\][/tex]
Thus, the correct choice is D.
