Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Trigonometric Functions in Terms of Sine and Cosine**
---
**Problem Statement:**
Write \( (\tan(x) + \sec(x)) (\sin(x) - 1) \) in terms of sine and cosine.
---
### Explanation:
This problem requires you to express the given trigonometric expression in terms of the basic trigonometric functions \( \sin(x) \) and \( \cos(x) \).
- **Tangent Function (\( \tan(x) \)):**
\[
\tan(x) = \frac{\sin(x)}{\cos(x)}
\]
- **Secant Function (\( \sec(x) \)):**
\[
\sec(x) = \frac{1}{\cos(x)}
\]
---
### Solution:
To rewrite the expression \( (\tan(x) + \sec(x)) (\sin(x) - 1) \) in terms of \( \sin(x) \) and \( \cos(x) \):
1. Substitute \( \tan(x) \) and \( \sec(x) \) with their sine and cosine forms:
\[
\tan(x) + \sec(x) = \frac{\sin(x)}{\cos(x)} + \frac{1}{\cos(x)}
\]
2. Combine the terms over a common denominator:
\[
\tan(x) + \sec(x) = \frac{\sin(x) + 1}{\cos(x)}
\]
3. Multiply by \( (\sin(x) - 1) \):
\[
(\tan(x) + \sec(x)) (\sin(x) - 1) = \left( \frac{\sin(x) + 1}{\cos(x)} \right) (\sin(x) - 1)
\]
4. Distribute \( (\sin(x) - 1) \):
\[
(\tan(x) + \sec(x)) (\sin(x) - 1) = \frac{(\sin(x) + 1)(\sin(x) - 1)}{\cos(x)}
\]
5. Use the difference of squares identity:
\[
(\sin(x) + 1)(\sin(x) - 1) = \sin^2(x) - 1
\]
Therefore,
\[
(\tan(x) + \sec(x)) (\sin](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58deed7b-e5cf-455c-8dcf-5c14f5ab5a29%2F443bf00e-92ae-4dd5-8a02-d42ea298b852%2Frty374f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Trigonometric Functions in Terms of Sine and Cosine**
---
**Problem Statement:**
Write \( (\tan(x) + \sec(x)) (\sin(x) - 1) \) in terms of sine and cosine.
---
### Explanation:
This problem requires you to express the given trigonometric expression in terms of the basic trigonometric functions \( \sin(x) \) and \( \cos(x) \).
- **Tangent Function (\( \tan(x) \)):**
\[
\tan(x) = \frac{\sin(x)}{\cos(x)}
\]
- **Secant Function (\( \sec(x) \)):**
\[
\sec(x) = \frac{1}{\cos(x)}
\]
---
### Solution:
To rewrite the expression \( (\tan(x) + \sec(x)) (\sin(x) - 1) \) in terms of \( \sin(x) \) and \( \cos(x) \):
1. Substitute \( \tan(x) \) and \( \sec(x) \) with their sine and cosine forms:
\[
\tan(x) + \sec(x) = \frac{\sin(x)}{\cos(x)} + \frac{1}{\cos(x)}
\]
2. Combine the terms over a common denominator:
\[
\tan(x) + \sec(x) = \frac{\sin(x) + 1}{\cos(x)}
\]
3. Multiply by \( (\sin(x) - 1) \):
\[
(\tan(x) + \sec(x)) (\sin(x) - 1) = \left( \frac{\sin(x) + 1}{\cos(x)} \right) (\sin(x) - 1)
\]
4. Distribute \( (\sin(x) - 1) \):
\[
(\tan(x) + \sec(x)) (\sin(x) - 1) = \frac{(\sin(x) + 1)(\sin(x) - 1)}{\cos(x)}
\]
5. Use the difference of squares identity:
\[
(\sin(x) + 1)(\sin(x) - 1) = \sin^2(x) - 1
\]
Therefore,
\[
(\tan(x) + \sec(x)) (\sin
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