Let's solve the problem step-by-step.
We are given that [tex]\(\tan \theta = -1\)[/tex] within the interval [tex]\(\frac{3\pi}{2} < \theta < 2\pi\)[/tex].
1. Identifying the Quadrant:
- The interval [tex]\(\frac{3\pi}{2} < \theta < 2\pi\)[/tex] corresponds to the fourth quadrant of the unit circle.
2. Behavior of [tex]\(\tan \theta\)[/tex] in the Fourth Quadrant:
- In the fourth quadrant, the tangent function ([tex]\(\tan \theta\)[/tex]) is negative. Thus, [tex]\(\tan \theta = -1\)[/tex] makes sense.
3. Determining the Angle:
- [tex]\(\tan \theta = -1\)[/tex] implies that [tex]\(\theta\)[/tex] corresponds to the reference angle where the tangent function equals 1 (but negative in the fourth quadrant).
- The angle where [tex]\(\tan \theta = 1\)[/tex] is [tex]\(\frac{\pi}{4}\)[/tex]. Therefore, in the fourth quadrant, the reference angle [tex]\(\theta\)[/tex] is [tex]\(2\pi - \frac{\pi}{4} = \frac{7\pi}{4}\)[/tex].
4. Finding [tex]\(\sec \theta\)[/tex]:
- [tex]\(\sec \theta\)[/tex] is the reciprocal of [tex]\(\cos \theta\)[/tex].
- Now, knowing that [tex]\(\theta = \frac{7\pi}{4}\)[/tex], we find [tex]\(\cos \left( \frac{7\pi}{4} \right)\)[/tex].
- The cosine of [tex]\(\frac{7\pi}{4}\)[/tex] is positive in the fourth quadrant.
- Specifically, [tex]\(\cos \left( \frac{7\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}\)[/tex].
- Thus, [tex]\(\sec \left( \frac{7\pi}{4} \right) = \frac{1}{\cos \left( \frac{7\pi}{4} \right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}\)[/tex].
Therefore, the value of [tex]\(\sec \theta\)[/tex] for [tex]\(\frac{3\pi}{2} < \theta < 2\pi\)[/tex] and given [tex]\(\tan \theta = -1\)[/tex] is [tex]\(\sqrt{2}\)[/tex].
So, the correct answer is [tex]\(\sqrt{2}\)[/tex].
