Problem 67
Question
Show that \((\sec \theta+1)(\sec \theta-1)=\tan ^{2} \theta\) is an identity.
Step-by-Step Solution
Verified Answer
By manipulating the left hand side of the given equation, using the reciprocal relationship of secant and cosine, and the Pythagorean identity, we have proven that indeed \((\sec \theta+1)(\sec \theta-1)=\tan ^{2} \theta \), showing this expression is an identity.
1Step 1:Multiply the left-hand side (LHS) of the equation
By using the property of difference of two squares for \((\sec \theta+1)(\sec \theta-1)\) , the expression simplifies to \( (\sec ^{2} \theta - 1) \).
2Step 2: Substitute the secant function
We can substitute \(\sec ^{2} \theta\) with \(1/\cos ^{2} \theta \). Thus, the expression becomes \((1/\cos ^{2} \theta - 1)\).
3Step 3: Rewrite the expression
Rewrite the expression with common denominator. It becomes \( \frac{1- \cos ^{2} \theta}{\cos ^{2} \theta} \).
4Step 4: Using the Pythagorean identity
We use the Pythagorean identity \(\sin ^{2} \theta + \cos ^{2} \theta = 1\) to replace \(1- \cos ^{2} \theta\) with \(\sin ^{2} \theta\). Thus, the expression becomes \( \frac{\sin ^{2} \theta}{\cos ^{2} \theta} \).
5Step 5: Simplify the expression
The ratio of sine to cosine is tangent: \(\tan \theta = \sin \theta / \cos \theta\). Thus, the expression becomes \( \tan ^{2} \theta\) which is equal to the right hand side (RHS).
Key Concepts
Difference of SquaresPythagorean IdentitySecant FunctionTangent Function
Difference of Squares
The difference of squares is a useful algebraic identity to know, especially for simplifying expressions like \((a + b)(a - b)\). This identity states that \((a + b)(a - b) = a^2 - b^2\).
Here's why it's helpful:
This step is crucial in connecting trigonometric identities, especially once we incorporate the Pythagorean Identity.
Here's why it's helpful:
- It allows complex terms to be simplified quickly.
- It's the foundation for many algebraic manipulations.
This step is crucial in connecting trigonometric identities, especially once we incorporate the Pythagorean Identity.
Pythagorean Identity
The Pythagorean Identity is one of the fundamental trigonometric identities. It states:\(\sin^2 \theta + \cos^2 \theta = 1\).
This relation is crucial for transforming and simplifying trigonometric expressions.
This relation is crucial for transforming and simplifying trigonometric expressions.
- By rearranging the identity, \(\cos^2 \theta = 1 - \sin^2 \theta\), we can substitute into other expressions.
- In this problem, the identity helps us express \(1 - \cos^2 \theta\) as \(\sin^2 \theta\), which further simplifies our expression.
Secant Function
The secant function, denoted as \(\sec \theta\), is one of the reciprocal trigonometric functions.Here's what you need to know:
Recognizing the secant function's role in these equations is critical to solving trigonometric identities effectively.
- It’s the reciprocal of the cosine function: \(\sec \theta = \frac{1}{\cos \theta}\).
- It helps express trigonometric functions in terms of sine and cosine, which are the more basic trigonometric functions.
Recognizing the secant function's role in these equations is critical to solving trigonometric identities effectively.
Tangent Function
The tangent function is another critical trigonometric function, represented by \(\tan \theta\).Here’s how it works:
At the end of the problem, the transformation \(\frac{\sin^2 \theta}{\cos^2 \theta}\) to \(\tan^2 \theta\) completes our demonstration. The goal was to show that the simplified version of the left side of the equation actually equaled \(\tan^2 \theta\), confirming that \((\sec \theta+1)(\sec \theta-1)=\tan ^{2} \theta\) is indeed a valid trigonometric identity.
- It represents the ratio of sine to cosine: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).
- Its square, \(\tan^2 \theta\), plays an essential role when simplifying expressions—like in this exercise.
At the end of the problem, the transformation \(\frac{\sin^2 \theta}{\cos^2 \theta}\) to \(\tan^2 \theta\) completes our demonstration. The goal was to show that the simplified version of the left side of the equation actually equaled \(\tan^2 \theta\), confirming that \((\sec \theta+1)(\sec \theta-1)=\tan ^{2} \theta\) is indeed a valid trigonometric identity.
Other exercises in this chapter
Problem 67
Graph each function in the interval from 0 to 2\(\pi\). \(y=-2 \cos \left(x-\frac{\pi}{3}\right)-4\)
View solution Problem 67
Solve each trigonometric equation for \(\theta\) in terms of \(y\) Sample $$\begin{aligned} y &=2 \sin 3 \theta+4 \\ \sin 3 \theta &=\frac{y-4}{2} \\\ 3 \theta
View solution Problem 68
In \(\triangle A B C, m \angle A=87^{\circ}, b=22 \mathrm{m},\) and \(c=19 \mathrm{m} .\) Find \(a\)
View solution Problem 68
Solve each trigonometric equation for \(\theta\) in terms of \(y\) Sample $$\begin{aligned} y &=2 \sin 3 \theta+4 \\ \sin 3 \theta &=\frac{y-4}{2} \\\ 3 \theta
View solution