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Chapter 6: Problem 23

Find the products and simplify your answers. $$(\csc \alpha-1)(\csc \alpha+1)$$

Short Answer

Expert verified
%\ (\csc \alpha - 1)(\csc \alpha + 1)\) simplifies to \(\cot^2 \alpha\).

Step by step solution

01

Identify the Expression

The given expression is \((\csc \alpha-1)(\csc \alpha+1)\). This is in the form of \((a - b)(a + b)\).
02

Recall the Formula for Difference of Squares

The formula for difference of squares states that \((a - b)(a + b) = a^2 - b^2\).
03

Apply the Formula to the Expression

We will use the difference of squares formula, setting \(a = \csc \alpha\) and \(b = 1\). The expression becomes \((\csc \alpha)^2 - 1^2\).
04

Simplify the Expression

Simplify \((\csc \alpha)^2 - 1^2\) to obtain \(\csc^2 \alpha - 1\).
05

Use the Pythagorean Identity

Recall that \(\csc^2 \alpha = 1 + \cot^2 \alpha\). Therefore, substituting this identity into the expression, we have: \(\csc^2 \alpha - 1 = (1 + \cot^2 \alpha) - 1\).
06

Final Simplification

Simplify the expression further to obtain \(\cot^2 \alpha\). This is the simplified form of the original expression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Squares
The difference of squares is a powerful algebraic identity used to simplify expressions of the form \(a^2 - b^2\). The formula is given by \((a - b)(a + b) = a^2 - b^2\). In our exercise, we started with the expression \( (\csc \alpha - 1)(\csc \alpha + 1) \). By recognizing this form, we identified it as a candidate for the difference of squares.
We set \a = \csc \alpha\ and \b = 1\, resulting in \( \csc^2 \alpha - 1 \). This simplification makes it easier to handle and prepares us to use other identities for further simplification.
Pythagorean Identity
The Pythagorean Identities are fundamental in trigonometry, relating the functions of sine, cosine, and their reciprocals. One of the key identities is: \csc^2 \alpha = 1 + \cot^2 \alpha\, which stems from the base identity \sin^2 \theta + \cos^2 \theta = 1\.
In our problem, we substituted \csc^2 \alpha\ with \(1 + \cot^2 \alpha \) into our simplified expression \( \csc^2 \alpha - 1 \). This step is crucial because it allows us to transform the expression into something more recognizable and simpler to work with.
The resulting expression is then \cot^2 \alpha\. Knowing and effectively applying these identities helps streamline the problem-solving process.
Trigonometric Function Simplification
Simplifying trigonometric functions involves using identities and algebraic manipulation to convert expressions into more manageable forms.
In our exercise, we began with the product \( (\csc \alpha - 1)(\csc \alpha + 1) \), identified it as a difference of squares, and reduced it to \( \csc^2 \alpha - 1 \).
Applying the Pythagorean Identity, we further simplified this to \( \cot^2 \alpha \). These steps reduced a complex trigonometric expression to a much simpler term, making it easier to understand and work with.
Always keep in mind a few common identities and algebraic techniques; they are invaluable tools for simplifying and solving trigonometric problems.

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Most popular questions from this chapter

Find all values of \(\theta\) in the interval \(0^{\circ}, 360^{\circ}\) ) that satisfy each \right. equation. Round approximate answers to the nearest tenth of a degree. $$\sec ^{4} \theta-5 \sec ^{2} \theta+4=0$$

Find all values of \(\alpha\) in degrees that satisfy each equation. Round approximate answers to the nearest tenth of a degree. $$\sec 2 \alpha=4.5$$

Find the central angle (to the nearest tenth of a degree) that intercepts an arc of length 5 feet on a circle of radius 60 feet.

In each case, find \(\sin \alpha, \cos \alpha, \tan \alpha, \csc \alpha, \sec \alpha,\) and \(\cot \alpha\) $$\sin (2 \alpha)=-8 / 17 \text { and } 180^{\circ}<2 \alpha<270^{\circ}$$

One of the acute angles of a right triangle is \(26^{\circ}\) and its hypotenuse is 38.6 inches. Find the lengths of its legs to the nearest tenth of an inch.
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