Problem 78
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. \( \dfrac{\tan x}{1 + \sec x} + \dfrac{1 + \sec x}{\tan x} \)
Step-by-Step Solution
Verified Answer
The simplified form of the expression \( \dfrac{\tan x}{1 + \sec x} + \dfrac{1 + \sec x}{\tan x} \) is \( \dfrac{2 + \cos x}{\sin x (1 + \cos x)} \)
1Step 1: Rewrite trigonometric expressions
Firstly, rewrite the expression using sine and cosine because they are the primary trigonometric functions and others can be expressed in terms of them. \nHere, \( \tan x = \dfrac{\sin x}{\cos x} \) and \( \sec x = \dfrac{1}{\cos x} \). So, substitute these values into the original expression: \( \dfrac{\dfrac{\sin x}{\cos x}}{1 + \dfrac{1}{\cos x}} + \dfrac{1 + \dfrac{1}{\cos x}}{\dfrac{\sin x}{\cos x}} \)
2Step 2: Simplify the fractions
The next step is to simplify the complex fractions. To do that, multiply the numerator and the denominator of each fraction by \( \cos x \): \[ \dfrac{\sin x}{1 + \cos x} + \dfrac{\cos x + 1}{\sin x} \] Now, making the denominators the same for both fractions by multiplying the first fraction by \( \sin x \) and the second by \( 1 + \cos x \) can help to simplify the expression further.
3Step 3: Perform addition operation
After step 2, the fractions' denominators are the same, allowing them to be easily added up. The result, after simplifying, will be: \[ \dfrac{\sin^2x + \cos x + 1}{\sin x (1 + \cos x)} \]
4Step 4: Use Pythagorean identity to simplify
The Pythagorean identity \( \sin^2x + \cos^2x = 1 \) has been used to simplify the expression. The result is: \[ \dfrac{2 + \cos x}{\sin x (1 + \cos x)} \]
Key Concepts
Sine and CosineTangent and SecantPythagorean IdentitySimplification of Trigonometric Expressions
Sine and Cosine
The sine and cosine functions are fundamental to understanding trigonometry. These functions relate to angles in right triangles. The sine of an angle, denoted as \( \sin x \), is defined as the ratio of the length of the opposite side to the length of the hypotenuse in a right triangle. Conversely, the cosine of an angle, denoted as \( \cos x \), is the ratio of the adjacent side to the hypotenuse.
In situations where more complex trigonometric functions like tangent or secant appear, it's often useful to express them in terms of sine and cosine. For instance:
In situations where more complex trigonometric functions like tangent or secant appear, it's often useful to express them in terms of sine and cosine. For instance:
- \( \tan x = \dfrac{\sin x}{\cos x} \)
- \( \sec x = \dfrac{1}{\cos x} \)
Tangent and Secant
Tangent and secant are two important trigonometric functions that arise often in various problems. The tangent function, \( \tan x \), represents the ratio between the sine and cosine of an angle. Meanwhile, the secant function, \( \sec x \), is the reciprocal of the cosine function; it shows how far away you are from the cosine value equal to zero.
Here's why these functions are valuable:
Here's why these functions are valuable:
- \( \tan x = \dfrac{\sin x}{\cos x} \), meaning when sine and cosine ratios are known, finding the tangent is straightforward.
- \( \sec x = \dfrac{1}{\cos x} \), allowing us to easily explore relationships involving the cosine function.
Pythagorean Identity
One of the most fundamental identities in trigonometry is the Pythagorean Identity. This identity involves the sine and cosine functions and states that for any angle \( x \), the following is always true: \( \sin^2 x + \cos^2 x = 1 \).
This identity is derived from the Pythagorean theorem in the context of the unit circle, where the radius of the circle is 1. The identity serves as a tool to simplify trigonometric expressions and verify other trigonometric identities. In our exercise, it allows us to replace \( \sin^2 x \) with \( 1 - \cos^2 x \), which can be incredibly useful when simplifying expressions.
This identity is derived from the Pythagorean theorem in the context of the unit circle, where the radius of the circle is 1. The identity serves as a tool to simplify trigonometric expressions and verify other trigonometric identities. In our exercise, it allows us to replace \( \sin^2 x \) with \( 1 - \cos^2 x \), which can be incredibly useful when simplifying expressions.
Simplification of Trigonometric Expressions
Simplification of trigonometric expressions is a crucial skill in trigonometry. It involves using trigonometric identities to express complex expressions in a simpler or more convenient form. The process often includes steps such as common denominator finding, applying identities, and algebraic manipulation.
In the given exercise, we dealt with a complicated expression involving \( \dfrac{\tan x}{1 + \sec x} + \dfrac{1 + \sec x}{\tan x} \). By expressing tangent and secant in terms of sine and cosine, and using the Pythagorean identity, we could effectively simplify the given expression:
In the given exercise, we dealt with a complicated expression involving \( \dfrac{\tan x}{1 + \sec x} + \dfrac{1 + \sec x}{\tan x} \). By expressing tangent and secant in terms of sine and cosine, and using the Pythagorean identity, we could effectively simplify the given expression:
- Substituting \( \tan x \) and \( \sec x \) using their sine and cosine forms.
- Finding a common denominator for the complex fraction addition.
- Simplifying the result using the identity: \( \sin^2 x + \cos^2 x = 1 \).
Other exercises in this chapter
Problem 78
In Exercises 75 - 78, use a graphing utility to approximate the solutions (to three decimal places) of the equation in the given interval. \( 2 \sec^2 x + \tan
View solution Problem 78
Write a short paper in your own words explaining to a classmate the difference between a trigonometric identity and a conditional equation. Include suggestions
View solution Problem 79
In Exercises 77-80, find all solutions of the equation in the interval \( [0, 2\pi) \). Use a graphing utility to graph the equation and verify the solutions. \
View solution Problem 79
In Exercises \(79-84,\) (a) use a graphing utility to graph the function and approximate the maximum and minimum points on the graph in the interval \([0,2 \pi)
View solution