Problem 38 - Contemporary Precalculus

Question

In Exercises \(37-42\), factor and simplify the given expression. $$\tan ^{2} t-\cot ^{2} t$$

Step-by-Step Solution

Answer

Question: Simplify and factor the expression \(\tan^2(t) - \cot^2(t)\). Answer: The simplified and factored expression is \(\sec^2(t) - \frac{\sec^2(t)}{\tan^2(t)}\).

Step 1: Recall trigonometric identities

Recall that \(\cot x = \frac{1}{\tan x}\), and then we have: $$ \tan ^{2}(t)-\cot ^{2}(t) = \tan ^{2}(t)- \left( \frac{1}{\tan(t)} \right) ^{2} $$

Step 2: Find common denominator

To simplify this expression, we need to find a common denominator for these terms. Since the denominator of the second term is \(\tan^2(t)\), we can rewrite the first term with the same denominator: $$ \frac{\tan^4(t)}{\tan^2(t)} - \frac{1}{\tan^2(t)} $$

Step 3: Combine terms under common denominator

Now that our terms have the same denominator, we can combine the numerators under one denominator: $$ \frac{\tan^4(t) - 1}{\tan^2(t)} $$

Step 4: Apply difference of squares factorization

We can see that the numerator is the difference of squares, so we can utilize the factorization \(\left(a^2 - b^2\right) = \left(a + b\right)\left(a - b\right)\). Let's apply this method: $$ \frac{(\tan^2(t) + 1)(\tan^2(t) - 1)}{\tan^2(t)} $$

Step 5: Use Pythagorean identity

Recall the Pythagorean identity that states: \(\tan^2(t) + 1 = \sec^2(t)\). Therefore, our expression becomes: $$ \frac{\sec^2(t)(\tan^2(t) - 1)}{\tan^2(t)} $$

Step 6: Simplify the expression

The expression can now be simplified as follows: $$ \boxed{\sec^2(t) - \frac{\sec^2(t)}{\tan^2(t)}} $$

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in simplifying expressions and solving equations in trigonometry. They are equations that relate the trigonometric functions to one another. For instance, in this exercise, knowing that

• \( an x = \frac{\sin x}{\cos x}\)
• \(\cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}\)

helps us transform the original expression effectively. To work with expressions involving \(\tan\) and \(\cot\), understanding they have an inverse relationship is crucial. This initial knowledge allows us to break down and reassemble the expression in more manageable parts.

Difference of Squares

The difference of squares is a powerful algebraic tool used to factor and simplify many expressions. The difference of squares formula is given by:

• \(a^2 - b^2 = (a + b)(a - b)\)

This pattern is applicable when you notice two perfect squares subtracted from each other. In the original problem, after obtaining the common denominator and combining the terms, the numerator becomes \(\tan^4(t) - 1\), which clearly fits the difference of squares form with \(a = \tan^2(t)\) and \(b = 1\). Recognizing and applying this pattern allows us to further simplify and transform the expression.

Pythagorean Identity

The Pythagorean identity is one of the most well-known and frequently used trigonometric identities:

• \(\tan^2(t) + 1 = \sec^2(t)\)

This identity stems from the Pythagorean theorem in relation to the unit circle and helps in transforming expressions by simplifying or substituting parts of the expression. In the context of the problem, applying this identity to \(\tan^2(t) + 1\) allows us to substitute into \(\sec^2(t)\), thereby achieving a simpler form. Using identities like this one can help streamline the process of simplifying trigonometric expressions.

Common Denominator in Algebra

Finding a common denominator is an essential step in simplifying expressions that contain fractions or rational terms. In this exercise, to combine terms

• \(\tan^2(t) - \cot^2(t)\)

we need a shared base, which involves rewriting the terms with a common denominator. The denominator here is \(\tan^2(t)\), and by rewriting each term accordingly, we form a single fraction. This method is not limited to trigonometry but is useful across various algebraic manipulations, making the task of addition or subtraction across fractions much simpler.