(cot A-csc A)(cosA+1)3D-sin A OSA+
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
Related questions
Question
Prove the identity
![### Trigonometric Identity
The following equation demonstrates a trigonometric identity that involves cotangent (cot), cosecant (csc), cosine (cos), and sine (sin) functions:
\[
(cot A - csc A)(cos A + 1) = -\sin A
\]
#### Explanation:
- **cot A**: Cotangent of angle A, which is the reciprocal of tangent (tan A) or \(\frac{1}{\tan A} = \frac{\cos A}{\sin A}\).
- **csc A**: Cosecant of angle A, which is the reciprocal of sine (sin A) or \(\frac{1}{\sin A}\).
- **cos A**: Cosine of angle A.
- **sin A**: Sine of angle A.
These types of identities are useful for simplifying trigonometric expressions and solving equations in various areas of mathematics, including algebra and calculus.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35a669ef-8ceb-4476-b34d-cf3aac0d1454%2Fe04c0336-81db-4020-826b-f67261b30016%2Fox88seo_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:### Trigonometric Identity
The following equation demonstrates a trigonometric identity that involves cotangent (cot), cosecant (csc), cosine (cos), and sine (sin) functions:
\[
(cot A - csc A)(cos A + 1) = -\sin A
\]
#### Explanation:
- **cot A**: Cotangent of angle A, which is the reciprocal of tangent (tan A) or \(\frac{1}{\tan A} = \frac{\cos A}{\sin A}\).
- **csc A**: Cosecant of angle A, which is the reciprocal of sine (sin A) or \(\frac{1}{\sin A}\).
- **cos A**: Cosine of angle A.
- **sin A**: Sine of angle A.
These types of identities are useful for simplifying trigonometric expressions and solving equations in various areas of mathematics, including algebra and calculus.
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