Problem 76 - Precalculus with Limits

Question

In Exercises 75 - 80, perform the addition or subtraction and use the fundamental identities to simplify. There is more than one correct form of each answer. \( \dfrac{1}{\sec x + 1} - \dfrac{1}{\sec x - 1} \)

Step-by-Step Solution

Answer

The final simplification is \( -2 \cot x \tan x \)

Step 1: Reduce to common denominator

To add or subtract fractions, we need the same denominator. Multiply the first fraction by \( \dfrac{\sec x - 1}{\sec x - 1} \) and the second fraction by \( \dfrac{\sec x + 1}{\sec x + 1} \), which are equivalent to 1.

Step 2: Simplification

This multiplication will give your algebraic expression the form \( \dfrac{a^2 - b^2}{c^2 - d^2} \), where \( a=\sec x -1, b=\sec x +1, c=\sec x - 1, d=\sec x +1 \). Now, \( a^2 - b^2 \) is a difference of squares which simplifies to \( (a-b)(a + b) \). Simplify the expression to get the final result.

Step 3: Apply Trig Identity

Substitute sec x with its reciprocal identity 1/cos x. This will simplify the expression further. Perform this substitution to find your final answer.

Key Concepts

Reciprocal Identities

In trigonometry, reciprocal identities are fundamental for simplifying expressions involving trigonometric functions. A reciprocal identity expresses a trigonometric function as the reciprocal of another. For example, the secant function is the reciprocal of the cosine function. So, we write \( \sec x = \frac{1}{\cos x} \). This is especially handy when simplifying trigonometric expressions or solving equations, as seen in the given problem where \( \sec x \) is converted to \( \frac{1}{\cos x} \).

• This substitution can make complex expressions more manageable.
• Always remember that every trigonometric function has a corresponding reciprocal identity.
• For example, \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{1}{\tan x} \).

Utilizing these identities helps in transforming and simplifying expressions which is a key step in solving problems related to trigonometric equations.

Difference of Squares

The difference of squares is a powerful algebraic concept commonly used to simplify polynomial expressions. It follows the pattern: \( a^2 - b^2 = (a - b)(a + b) \). This principle allows us to handle the expression more easily by breaking it into a product of two binomials.
In our problem, after forming a common denominator for the fractions, the expression can be arranged into the form \( a^2 - b^2 \). Here, \( a = (\sec x - 1) \) and \( b = (\sec x + 1) \). Applying the difference of squares, we get \( (\sec x - 1)^2 - (\sec x + 1)^2 = ((\sec x - 1) - (\sec x + 1))((\sec x - 1) + (\sec x + 1)) \).

• This step is crucial for simplicity as it reduces complex expressions to more manageable forms.
• Understanding the difference of squares formula helps quickly identify and simplify these patterns.

Algebraic Expressions

Algebraic expressions form the backbone of many mathematical applications, including trigonometry. They comprise numbers, variables, and operational symbols like \(+, -, *, \div\). In this exercise, we manipulate algebraic expressions to perform arithmetic with fractions and simplify the equation.
The compilation of these expressions requires a systematic approach:

• First, combine fractions using a common denominator.
• Apply algebraic patterns like the difference of squares.
• Use trigonometric identities to substitute and simplify further.

Understanding how to work with algebraic expressions is essential for most mathematical manipulations. It helps us reorganize and resolve expressions to find a meaningful or simplified form. This systematic approach is critical for solving more complex mathematical problems efficiently.