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...
Related questions
Question
Can you please show the steps
![## 1. Prove Trigonometric Identity
(a) \[ \cos\theta \, (\sec\theta - \cos\theta) = \sin^2\theta \]
(b) \[ \frac{\cos\theta}{1 - \sin\theta} = \sec\theta + \tan\theta \]
(c) \[ \frac{1 + \cos\theta}{\cos\theta} = \frac{\tan^2\theta}{\sec\theta - 1} \]
**Explanation:**
In this task, you are required to prove each of the given trigonometric identities. Here’s a breakdown of each statement:
1. **Part (a)**
- Starting expression: \(\cos\theta \, (\sec\theta - \cos\theta)\)
- You need to simplify this expression and show that it equals \(\sin^2\theta\).
2. **Part (b)**
- Starting expression: \(\frac{\cos\theta}{1 - \sin\theta}\)
- You need to show that this fraction simplifies to \(\sec\theta + \tan\theta\).
3. **Part (c)**
- Starting expression: \(\frac{1 + \cos\theta}{\cos\theta}\)
- You need to show that this fraction equals \(\frac{\tan^2\theta}{\sec\theta - 1}\).
Each of these problems can be solved using fundamental trigonometric identities, such as:
- Pythagorean identities:
- \(\sin^2\theta + \cos^2\theta = 1\)
- \(1 + \tan^2\theta = \sec^2\theta\)
- Reciprocal identities:
- \(\sec\theta = \frac{1}{\cos\theta}\)
- \(\csc\theta = \frac{1}{\sin\theta}\)
- Quotient identities:
- \(\tan\theta = \frac{\sin\theta}{\cos\theta}\)
- \(\cot\theta = \frac{\cos\theta}{\sin\theta}\)
By applying these identities, you can simplify and manipulate the given expressions to prove that they hold true.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedc716be-e066-4234-bd1a-1a86837575c3%2F7dfb056f-0e38-4b7f-b0c2-f752ba38aef1%2Fs1m73nk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## 1. Prove Trigonometric Identity
(a) \[ \cos\theta \, (\sec\theta - \cos\theta) = \sin^2\theta \]
(b) \[ \frac{\cos\theta}{1 - \sin\theta} = \sec\theta + \tan\theta \]
(c) \[ \frac{1 + \cos\theta}{\cos\theta} = \frac{\tan^2\theta}{\sec\theta - 1} \]
**Explanation:**
In this task, you are required to prove each of the given trigonometric identities. Here’s a breakdown of each statement:
1. **Part (a)**
- Starting expression: \(\cos\theta \, (\sec\theta - \cos\theta)\)
- You need to simplify this expression and show that it equals \(\sin^2\theta\).
2. **Part (b)**
- Starting expression: \(\frac{\cos\theta}{1 - \sin\theta}\)
- You need to show that this fraction simplifies to \(\sec\theta + \tan\theta\).
3. **Part (c)**
- Starting expression: \(\frac{1 + \cos\theta}{\cos\theta}\)
- You need to show that this fraction equals \(\frac{\tan^2\theta}{\sec\theta - 1}\).
Each of these problems can be solved using fundamental trigonometric identities, such as:
- Pythagorean identities:
- \(\sin^2\theta + \cos^2\theta = 1\)
- \(1 + \tan^2\theta = \sec^2\theta\)
- Reciprocal identities:
- \(\sec\theta = \frac{1}{\cos\theta}\)
- \(\csc\theta = \frac{1}{\sin\theta}\)
- Quotient identities:
- \(\tan\theta = \frac{\sin\theta}{\cos\theta}\)
- \(\cot\theta = \frac{\cos\theta}{\sin\theta}\)
By applying these identities, you can simplify and manipulate the given expressions to prove that they hold true.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps with 4 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning