Chapter 6: Problem 16
Show algebraically that \(\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}=2 \sec ^{2} \theta\) is an identity.
Short Answer
Expert verified
It is shown that \[\frac{1}{1+\text{sin}\theta}+\frac{1}{1-\text{sin}\theta} = 2\text{sec}^2\theta.\]
Step by step solution
01
- Express in a common denominator
Rewrite the expression \(\frac{1}{1+\text{sin}\theta} + \frac{1}{1-\text{sin}\theta}\) with a common denominator. The common denominator for these fractions is \( (1+\text{sin}\theta)(1-\text{sin}\theta) \).
02
- Combine under the common denominator
Combine the fractions: \[ \frac{(1-\text{sin}\theta)+(1+\text{sin}\theta)}{(1+\text{sin}\theta)(1-\text{sin}\theta)}. \]
03
- Simplify the numerator
Simplify the numerator: \[(1 - \text{sin}\theta) + (1 + \text{sin}\theta) = 2.\]
04
- Simplify the denominator
Simplify the denominator: \[(1+\text{sin}\theta)(1-\text{sin}\theta) = 1 - \text{sin}^2\theta \].
05
- Use Pythagorean identity
Recall the Pythagorean identity \( 1 - \text{sin}^2\theta = \text{cos}^2\theta \). Substituting this in the denominator, we get \[\frac{2}{\text{cos}^2\theta}.\]
06
- Result in terms of secant
Recognize that \( \text{sec}\theta = \frac{1}{\text{cos}\theta} \). Thus, \[\frac{2}{\text{cos}^2\theta} = 2\text{sec}^2\theta\].
07
- Conclude the identity proof
Thus, it can be concluded that \[\frac{1}{1+\text{sin}\theta}+\frac{1}{1-\text{sin}\theta} = 2\text{sec}^2\theta \].
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Common Denominator
When faced with adding or subtracting fractions, finding a common denominator ensures the process goes smoothly. To start, identify the denominators of the fractions you are working with. For our exercise, the fractions are \(\frac{1}{1+\text{sin}\theta}\) and \(\frac{1}{1-\text{sin}\theta}\). The common denominator is the product of the two different denominators: \((1+\text{sin}\theta)(1-\text{sin}\theta)\). This step is crucial for combining the fractions. By rewriting them with a common denominator, we are able to add or subtract them accordingly.
- The product rule of denominators helps in achieving a common base.
- Having a common denominator simplifies the overall expression for the next steps.
Pythagorean Identity
The Pythagorean identity is a fundamental relation in trigonometry that connects the sine and cosine functions. The identity is: \[ 1 - \text{sin}^2\theta = \text{cos}^2\theta \]. This comes in handy when simplifying trigonometric expressions. In our exercise, after expressing the fractions with a common denominator, we noticed that the denominator simplified to \1 - \text{sin}^2\theta\. By applying the Pythagorean identity, we replaced this with \ \text{cos}^2\theta \.
This identity not only simplifies calculations but also helps in transforming trigonometric expressions into more familiar or workable forms.
This identity not only simplifies calculations but also helps in transforming trigonometric expressions into more familiar or workable forms.
- The Pythagorean identity simplifies the proofs involving sin and cos.
- It helps in transforming products of trigonometric functions.
Secant Function
The secant function, denoted as \ \text{sec} \theta \, is the reciprocal of the cosine function. Mathematically, it is defined as: \[ \text{sec}\theta = \frac{1}{\text{cos}\theta} \]. In our exercise, once we simplified the expression to \ \frac{2}{\text{cos}^2\theta} \, recognizing the secant function was key to tying everything together. By knowing that \ \text{sec}\theta = \frac{1}{\text{cos}\theta} \, it follows that \ \frac{2}{\text{cos}^2\theta} \ simplifies to \ 2\text{sec}^2\theta \.
- The secant function is often used in identities and simplifications.
- It is a critical function for hyperbolic and inverse trigonometric relationships.
Algebraic Manipulation
Algebraic manipulation involves rearranging, simplifying and solving expressions using algebraic rules. For our problem, it included:
Mastering algebraic manipulation comes down to practice. Step by step, one can simplify and transform complex expressions into forms where known identities or values can be easily applied.
- Decomposing and distributing terms.
- Combining like terms.
- Using known identities to simplify expressions.
Mastering algebraic manipulation comes down to practice. Step by step, one can simplify and transform complex expressions into forms where known identities or values can be easily applied.
- Always proceed methodically.
- Keep track of each step.
- Verify by back substituting.