Problem 100

Question

Rewrite the expression as a single logarithm and simplify the result. (Hint: Begin by using the properties of logarithms.) $$\ln |\cot t|+\ln \left(1+\tan ^{2} t\right)$$

Step-by-Step Solution

Verified
Answer
\(\ln |\csc^2(t)|\), for \(t \neq n\pi\) where \(n\) is an integer
1Step 1: Apply the logarithm product rule
Using the properties of logarithms, \(\ln(a) + \ln(b) = \ln(a \cdot b)\), the expression becomes: \( \ln |\cot(t)(1 + \tan^2(t))|\).
2Step 2: Simplify the trigonometric expression
Recognizing that \(1 + \tan^2(t)\) is equal to \( \sec^2(t)\) (by the Pythagorean identity), the expression becomes: \(\ln |\cot(t)\sec^2(t)|\).
3Step 3: Convert \(\cot(t)\) and \(\sec^2(t)\) into sine and cosine
Rewriting \(\cot(t)\) as \( \cos(t)/\sin(t) \), and \(\sec^2(t) \) as \( 1/\cos^2(t) \), the expression becomes: \(\ln |(\cos(t)/\sin(t)) \cdot (1/\cos^2(t))|\).
4Step 4: Simplify the expression
Further simplify the expression within the absolute brackets, \(\cos(t)/\sin^2(t)\cos^2(t)\), becomes \(1/\sin^2(t)\) which equals \( \csc^2(t)\). Therefore, the expression becomes: \(\ln |\csc^2(t)|\). This expression is valid under the constraint that \(\sin(t) \neq 0\).

Key Concepts

Understanding the Logarithm Product RuleLeveraging Trigonometric Identities to Simplify ExpressionsSimplifying Logarithmic Expressions with Trigonometric Functions
Understanding the Logarithm Product Rule
The logarithm product rule is a fundamental property of logarithms that states when two numbers are multiplied together, the logarithm of the product is the sum of the logarithms of the individual numbers. Symbolically, this is represented as: \[ \ln(a \times b) = \ln(a) + \ln(b) \].
This property is incredibly useful when simplifying complex logarithmic expressions. In the given exercise, the product rule is applied to combine two logarithms into one. This step is pivotal as it sets the stage for further simplification of the expression.
Once you understand that the logarithm of a product is the sum of the logarithms, you can reverse this process to combine multiple log terms into a single one, which is the primary goal in simplifying logarithmic expressions.
Leveraging Trigonometric Identities to Simplify Expressions
Trigonometric identities are equations that are true for all values of the variable where both sides of the equation are defined. These identities are invaluable tools for simplifying trigonometric expressions and solving equations. The Pythagorean identities, for instance, express fundamental relationships among the trigonometric functions sine, cosine, and tangent.
One such identity is: \[ 1 + \tan^2(t) = \sec^2(t) \].
This identity is used in the exercise to rewrite the expression \( 1 + \tan^2(t) \) as \( \sec^2(t) \). Recognizing and applying these identities correctly can make complex expressions much simpler and more manageable. By breaking down trigonometric functions into their sine and cosine components, a student can more easily see pathways to simplification that might not have been apparent at first glance.
Simplifying Logarithmic Expressions with Trigonometric Functions
Simplifying logarithmic expressions that involve trigonometric functions can be challenging, but by diligently applying the properties of logarithms and utilizing trigonometric identities, the process becomes more systematic and easier to manage. In our exercise, after applying the logarithm product rule, it was critical to find a way to further simplify the trigonometric expression inside the logarithm.
This involved converting less common trigonometric functions like cotangent and secant into their more fundamental forms, sine and cosine. The expression: \[ \cot(t) \sec^2(t) \], becomes much more approachable when rewritten as: \[ \frac{\cos(t)}{\sin(t)} \times \frac{1}{\cos^2(t)} \].
Further simplification leads to: \[ \csc^2(t) \], thereby making the expression both simpler and neater. Throughout this process, keeping track of the domains where the functions are defined is crucial, as logarithms are not defined for non-positive numbers.