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Ch. 3 - Trigonometric Identities and Equations
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Not the one you use?Change textbook
Chapter 3, Problem 22

In Exercises 1–60, verify each identity. cot² t /csc t = csc t - sin t

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1
Start with the left-hand side (LHS) of the identity: \( \frac{\cot^{2} t}{\csc t} \). Recall the definitions of cotangent and cosecant in terms of sine and cosine: \( \cot t = \frac{\cos t}{\sin t} \) and \( \csc t = \frac{1}{\sin t} \).
Rewrite \( \cot^{2} t \) as \( \left( \frac{\cos t}{\sin t} \right)^{2} = \frac{\cos^{2} t}{\sin^{2} t} \). Substitute this and \( \csc t = \frac{1}{\sin t} \) into the LHS to get \( \frac{\frac{\cos^{2} t}{\sin^{2} t}}{\frac{1}{\sin t}} \).
Simplify the complex fraction by multiplying numerator and denominator: \( \frac{\cos^{2} t}{\sin^{2} t} \times \sin t = \frac{\cos^{2} t \cdot \sin t}{\sin^{2} t} = \frac{\cos^{2} t}{\sin t} \).
Now, focus on the right-hand side (RHS): \( \csc t - \sin t = \frac{1}{\sin t} - \sin t \). To combine these terms, write \( \sin t \) as \( \frac{\sin^{2} t}{\sin t} \) to get a common denominator: \( \frac{1}{\sin t} - \frac{\sin^{2} t}{\sin t} = \frac{1 - \sin^{2} t}{\sin t} \).
Use the Pythagorean identity \( 1 - \sin^{2} t = \cos^{2} t \) to rewrite the numerator, so the RHS becomes \( \frac{\cos^{2} t}{\sin t} \), which matches the simplified LHS, thus verifying the identity.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Trigonometric Identities

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