Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![### Trigonometric Problem
#### Problem Statement:
If \( \sec(\theta) = -3 \), and \( \theta \) is in the second quadrant, find \( \tan(\theta) \).
#### Multiple Choice Answers:
- \( \sqrt{10} \)
- \( 2\sqrt{2} \)
- \( -\sqrt{10} \)
- \( -2\sqrt{2} \)
To solve this problem, follow these steps:
1. **Recall the Definition of Secant:**
\[
\sec(\theta) = \frac{1}{\cos(\theta)}
\]
2. **Isolate \( \cos(\theta) \):**
\[
\cos(\theta) = \frac{1}{\sec(\theta)} = \frac{1}{-3} = -\frac{1}{3}
\]
3. **Use the Pythagorean Identity to Find \( \sin(\theta) \):**
\[
\cos^2(\theta) + \sin^2(\theta) = 1
\]
\[
\left(-\frac{1}{3}\right)^2 + \sin^2(\theta) = 1
\]
\[
\frac{1}{9} + \sin^2(\theta) = 1
\]
\[
\sin^2(\theta) = 1 - \frac{1}{9}
\]
\[
\sin^2(\theta) = \frac{9}{9} - \frac{1}{9}
\]
\[
\sin^2(\theta) = \frac{8}{9}
\]
\[
\sin(\theta) = \pm \sqrt{\frac{8}{9}} = \pm \frac{\sqrt{8}}{3} = \pm \frac{2\sqrt{2}}{3}
\]
Since \( \theta \) is in the second quadrant, \( \sin(\theta) \) must be positive:
\[
\sin(\theta) = \frac{2\sqrt{2}}{3}
\]
4. **Compute \( \tan(\theta) \):**
\[
\tan(\theta) = \frac{\sin](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4c67c4d0-217c-4e3e-983c-7902447cdce8%2Fd5a5f340-7ec1-4a26-91f9-85330e0a8ba2%2Fhfmuau8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Trigonometric Problem
#### Problem Statement:
If \( \sec(\theta) = -3 \), and \( \theta \) is in the second quadrant, find \( \tan(\theta) \).
#### Multiple Choice Answers:
- \( \sqrt{10} \)
- \( 2\sqrt{2} \)
- \( -\sqrt{10} \)
- \( -2\sqrt{2} \)
To solve this problem, follow these steps:
1. **Recall the Definition of Secant:**
\[
\sec(\theta) = \frac{1}{\cos(\theta)}
\]
2. **Isolate \( \cos(\theta) \):**
\[
\cos(\theta) = \frac{1}{\sec(\theta)} = \frac{1}{-3} = -\frac{1}{3}
\]
3. **Use the Pythagorean Identity to Find \( \sin(\theta) \):**
\[
\cos^2(\theta) + \sin^2(\theta) = 1
\]
\[
\left(-\frac{1}{3}\right)^2 + \sin^2(\theta) = 1
\]
\[
\frac{1}{9} + \sin^2(\theta) = 1
\]
\[
\sin^2(\theta) = 1 - \frac{1}{9}
\]
\[
\sin^2(\theta) = \frac{9}{9} - \frac{1}{9}
\]
\[
\sin^2(\theta) = \frac{8}{9}
\]
\[
\sin(\theta) = \pm \sqrt{\frac{8}{9}} = \pm \frac{\sqrt{8}}{3} = \pm \frac{2\sqrt{2}}{3}
\]
Since \( \theta \) is in the second quadrant, \( \sin(\theta) \) must be positive:
\[
\sin(\theta) = \frac{2\sqrt{2}}{3}
\]
4. **Compute \( \tan(\theta) \):**
\[
\tan(\theta) = \frac{\sin
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