To determine in which intervals the trigonometric inequality [tex]\(\sec(x) < \cot(x)\)[/tex] holds true, let's analyze the given options one by one.
### Given Options:
1. [tex]\(0 < x < \frac{\pi}{2}\)[/tex]
2. [tex]\(\frac{\pi}{2} < x < \pi\)[/tex]
3. [tex]\(\pi < x < \frac{3\pi}{2}\)[/tex]
4. [tex]\(\frac{3 \pi}{2} < x < 2 \pi\)[/tex]
To determine where [tex]\(\sec(x) < \cot(x)\)[/tex] is satisfied, let's consider each interval and see if the inequality consistently holds.
### Analyzing the Intervals
1. Interval [tex]\(0 < x < \frac{\pi}{2}\)[/tex]:
- In this interval, [tex]\( \sec(x)\)[/tex] is positive (since [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex] and [tex]\(\cos(x) > 0\)[/tex]).
- Also, [tex]\(\cot(x) = \frac{\cos(x)}{\sin(x)}\)[/tex] is positive (since both [tex]\(\cos(x)\)[/tex] and [tex]\(\sin(x)\)[/tex] are positive).
- As [tex]\(x\)[/tex] approaches 0, [tex]\(\sec(x)\)[/tex] approaches 1 (since [tex]\(\cos(x) \to 1\)[/tex]), and [tex]\(\cot(x)\)[/tex] approaches infinity (since [tex]\(\sin(x) \to 0\)[/tex]).
- As [tex]\(x\)[/tex] approaches [tex]\(\frac{\pi}{2}\)[/tex], [tex]\(\sec(x)\)[/tex] approaches infinity (since [tex]\(\cos(x) \to 0\)[/tex]), and [tex]\(\cot(x)\)[/tex] approaches 0 (since [tex]\(\sin(x) \to 1\)[/tex]).
- Therefore, [tex]\(\sec(x) < \cot(x)\)[/tex] is not consistently true across the whole interval.
2. Interval [tex]\(\frac{\pi}{2} < x < \pi\)[/tex]:
- For [tex]\(\frac{\pi}{2} < x < \pi\)[/tex], [tex]\(\cos(x)\)[/tex] is negative, making [tex]\(\sec(x)\)[/tex] negative.
- [tex]\(\sin(x)\)[/tex] is positive, making [tex]\(\cot(x) = \frac{\cos(x)}{\sin(x)}\)[/tex] negative as well.
- The magnitudes of [tex]\(\sec(x)\)[/tex] and [tex]\(\cot(x)\)[/tex] need to be compared. Here, [tex]\(\sec(x)\)[/tex] is less than [tex]\(\cot(x)\)[/tex] for the specified interval values.
3. Interval [tex]\(\pi < x < \frac{3\pi}{2}\)[/tex]:
- In this interval, both [tex]\(\cos(x)\)[/tex] and [tex]\(\sin(x)\)[/tex] are negative.
- The secant function [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex] will be negative.
- The cotangent function [tex]\(\cot(x) = \frac{\cos(x)}{\sin(x)}\)[/tex] is positive (since both [tex]\(\cos(x)\)[/tex] and [tex]\(\sin(x)\)[/tex] are negative, so their ratio is positive).
- Since one is negative and the other positive, [tex]\(\sec(x) < \cot(x)\)[/tex] holds true in this interval.
4. Interval [tex]\(\frac{3 \pi}{2} < x < 2 \pi\)[/tex]:
- In this interval, [tex]\(\cos(x)\)[/tex] is positive and [tex]\(\sin(x)\)[/tex] is negative.
- Hence, [tex]\(\sec(x)\)[/tex] is positive and [tex]\(\cot(x) = \frac{\cos(x)}{\sin(x)}\)[/tex] is negative.
- Since [tex]\(\one is positive and the other is negative, \(\sec(x) < \cot(x)\)[/tex] holds true in this interval.
### Conclusion
After examining each interval, the inequality [tex]\(\sec(x) < \cot(x)\)[/tex] always holds true in multiple intervals:
- [tex]\( \frac{\pi}{2} < x < \pi \)[/tex]
- [tex]\(\pi < x < \frac{3\pi}{2}\)[/tex]
Therefore, the intervals where [tex]\(\sec(x) < \cot(x)\)[/tex] consistently holds true are:
- [tex]\(\frac{\pi}{2} < x < \pi\)[/tex]
- [tex]\(\pi < x < \frac{3\pi}{2}\)[/tex]
- [tex]\(\frac{3\pi}{2} < x < 2\pi\)[/tex]
So, the correct intervals are:
[tex]\[ (\frac{\pi}{2}, \pi) \text{ radians and } (\pi, \frac{3\pi}{2}) \text{ radians.} \][/tex]
