Problem 44 - Algebra 2

Question

REVIEW Which of the following is equivalent to \(\frac{\cos \theta\left(\cot ^{2} \theta+1\right)}{\csc \theta} ?\) $$ \begin{array}{l}{\mathrm{F} \tan \theta} \\ {\mathrm{G} \cot \theta} \\\ {\mathrm{H} \sec \theta} \\ {\mathrm{J} \csc \theta}\end{array} $$

Step-by-Step Solution

Answer

The expression simplifies to \( \tan \theta \), corresponding to option F.

Step 1: Rewrite Trigonometric Functions

The given expression is \( \frac{\cos \theta (\cot^{2} \theta + 1)}{\csc \theta} \). Start by rewriting each trigonometric function in terms of sine and cosine. \( \cot \theta \) is \( \frac{\cos \theta}{\sin \theta} \), so \( \cot^2 \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \). Also, \( \csc \theta = \frac{1}{\sin \theta} \).

Step 2: Simplify \( \cot^2 \theta + 1 \) Using Pythagorean Identity

The identity \( \cot^2 \theta + 1 = \csc^2 \theta \) simplifies this part. Since \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \), the expression becomes \( \frac{\cos \theta \csc^2 \theta}{\csc \theta} \).

Step 3: Simplify the Expression

Replace \( \csc^2 \theta \) with \( \frac{1}{\sin^2 \theta} \) and \( \csc \theta \) with \( \frac{1}{\sin \theta} \). The expression now is \( \frac{\cos \theta \cdot \frac{1}{\sin^2 \theta}}{\frac{1}{\sin \theta}} \). Simplify by canceling \( \sin \theta \) in the numerator and denominator, which yields \( \cos \theta \cdot \sin \theta \).

Step 4: Identify Equivalent Function

The expression \( \cos \theta \cdot \sin \theta \) can be rewritten as \( \frac{\sin \theta}{\cos \theta} \cdot \cos^2 \theta = \sin \theta \cdot \cos \theta \), which results in \( \tan \theta \). Thus, the equivalent choice is \( \mathrm{F} \tan \theta \).

Key Concepts

Simplifying Trigonometric Expressions

Simplifying trigonometric expressions involves transforming them into a more manageable or recognizable form, often by using equivalent functions or identities. Let's explore how this works using basic trigonometric functions such as sine, cosine, and tangent.
When you encounter an expression like \( \frac{\cos \theta (\cot^{2} \theta + 1)}{\csc \theta} \), the goal is to reduce it to something simpler, like another standard trigonometric function. This usually begins by rewriting the involved functions in terms of sine and cosine because these are the basic building blocks of many transformations.

• Start by recalling some basic conversions: \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \). Rewriting functions in this way allows us to see how different parts of an expression interact and can often reveal the simplest form of the expression.
• The simplification process might also involve identities, which can replace more complex sections of an expression with something more straightforward.

The result should be a simpler version that either expresses the original idea more clearly or allows for further simplification steps to take place.

Pythagorean Identities

The Pythagorean identities are essential tools in simplifying expressions and solving equations involving trigonometric functions. They stem from the Pythagorean Theorem and describe relationships between them.
One fundamental identity is \( \sin^2 \theta + \cos^2 \theta = 1 \). There are other variations involving tangent and cotangent functions, such as \( 1 + \cot^2 \theta = \csc^2 \theta \), which was used in the step-by-step solution of our exercise.

• When faced with \( \cot^2 \theta + 1 \), recognized as a Pythagorean identity, it immediately simplifies to \( \csc^2 \theta \).
• These identities allow us to consolidate terms and find equivalences that are less obvious initially. For instance, simplifying expressions like \( \cot^2 \theta + 1 \) becomes straightforward by recognizing it as \( \csc^2 \theta \).

Mastering these identities is crucial as they often simplify what would otherwise be cumbersome algebraic tasks.

Trigonometric Functions

Understanding trigonometric functions is fundamental in working with expressions and equations in trigonometry. These functions describe the relationships between the angles and sides of triangles.
The core functions include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)), which are often expressed in terms of a right triangle's sides:

• \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \)
• \( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \)
• \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \)

Other functions like cotangent (\( \cot \)), secant (\( \sec \)), and cosecant (\( \csc \)) are reciprocals of the primary functions:

• \( \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \)
• \( \sec \theta = \frac{1}{\cos \theta} \)
• \( \csc \theta = \frac{1}{\sin \theta} \)

Having a strong grip on these fundamental functions is essential for solving problems and simplifying expressions in trigonometry. It allows us to switch effortlessly between different forms and recognize when complex or compound expressions can be reduced to simpler known forms.