Problem 101
Question
Verify that each equation is an identity by using any of the identities introduced in the first three sections of this chapter. $$\frac{\tan ^{2} t+1}{\tan t \csc ^{2} t}=\tan t$$
Step-by-Step Solution
Verified Answer
The equation is verified as an identity.
1Step 1: Recall Identity Relationships
The identity \( \tan^2 t + 1 = \sec^2 t \) is one of the Pythagorean identities.
2Step 2: Substitute the Pythagorean Identity
Substitute \( \tan^2 t + 1 \) with \( \sec^2 t \) in the left-hand side of the equation: \[ \frac{\sec^2 t}{\tan t \csc^2 t} \]
3Step 3: Rewrite Trigonometric Functions
Rewrite \( \sec^2 t \) and \( \csc^2 t \) in terms of sine and cosine:\[ \sec^2 t = \frac{1}{\cos^2 t}, \quad \csc^2 t = \frac{1}{\sin^2 t} \] Substitute these into the equation: \[ \frac{\frac{1}{\cos^2 t}}{\tan t \cdot \frac{1}{\sin^2 t}} \]
4Step 4: Simplify the Denominator
We know \( \tan t = \frac{\sin t}{\cos t} \). Thus: \[ \tan t \cdot \frac{1}{\sin^2 t} = \frac{\sin t}{\cos t} \cdot \frac{1}{\sin^2 t} = \frac{1}{\sin t \cos t} \]
5Step 5: Simplify the Entire Expression
Substitute and simplify the entire expression:\[ \frac{\frac{1}{\cos^2 t}}{\frac{1}{\sin t \cos t}} = \frac{1}{\cos^2 t} \times \frac{\sin t \cos t}{1} = \frac{\sin t}{\cos t} = \tan t \]The left-hand side simplifies to \( \tan t \), which matches the right-hand side.
Key Concepts
Understanding Pythagorean IdentitiesTrigonometric Simplification TechniquesBasic Trigonometric Functions
Understanding Pythagorean Identities
Pythagorean identities are fundamental relationships in trigonometry that connect different trigonometric functions. They stem from the Pythagorean theorem, which applies to right-angled triangles. The most common Pythagorean identity is \( \sin^2 \theta + \cos^2 \theta = 1 \). From this base identity, we can derive others, such as:
- \( \tan^2 \theta + 1 = \sec^2 \theta \)
- \( 1 + \cot^2 \theta = \csc^2 \theta \)
Trigonometric Simplification Techniques
Trigonometric simplification involves reducing complex trigonometric expressions into simpler forms. This process uses various identities and algebraic manipulations to make calculations manageable. In the given solution, we simplify the original expression \( \frac{\tan^2 t+1}{\tan t \csc^2 t} \) by first substituting the Pythagorean identity \( \tan^2 t + 1 = \sec^2 t \). This step simplifies the expression to \( \frac{\sec^2 t}{\tan t \csc^2 t} \). The next step uses basic trigonometric conversions, rewriting \( \sec^2 t \) and \( \csc^2 t \) in terms of \( \cos \) and \( \sin \), allowing for further simplifications. We also simplify the denominator by expressing \( \tan t \) in terms of \( \sin t \) and \( \cos t \), leading to a cancellation of terms and ultimately simplifying the expression to match the desired reminder. By understanding how to swap more complex elements for their simpler equivalents, you can effortlessly handle trigonometric expressions. Incorporating identities into simplification will become a powerful ally.
Basic Trigonometric Functions
Trigonometric functions, such as sine, cosine, and tangent, are foundational mathematical functions that describe relationships in right-angled triangles and model periodic phenomena like waves. Each function has reciprocal counterparts:
- \( \sin \theta \) (sine) becomes \( \csc \theta \) (cosecant)
- \( \cos \theta \) (cosine) becomes \( \sec \theta \) (secant)
- \( \tan \theta \) (tangent) becomes \( \cot \theta \) (cotangent)
Other exercises in this chapter
Problem 100
Use a calculator to find each value. $$\cot (\arccos 0.58236841)$$
View solution Problem 100
Verify that equation is an identity. \(\frac{\tan s}{1+\cos s}+\frac{\sin s}{1-\cos s}=\cot s+\sec s \csc s\)
View solution Problem 101
Write each expression as an algebraic expression in \(u, u>0\). $$\sec \left(\cos ^{-1} u\right)$$
View solution Problem 101
Verify that equation is an identity. \(\frac{\tan ^{2} t-1}{\sec ^{2} t}=\frac{\tan t-\cot t}{\tan t+\cot t}\)
View solution