Establish the identity. cos (α - B) cos a cos ß Choose the sequence of steps below that verifies the identity. sin a sin - cos acos B cos a cos O A. O B. O C. = 1+ tan atan ß O D. cos (α - B) cos a cos B cos (α - B) cos a cos B cos (α - B) cos a cos B cos (α - B) cos x cos ß cos a cos cos sin a sin B cos a cos ß cos acos - sin a sin ß cos acos ß acos B + sin asin ß cos a cos sin a sin B sin a sin B + cos a cos ß sin a sin ß cos a cos cos a cos cos acos ß+ sin asin ß cos a cos ß + cos a cos ß cos a cos ß cos acos B sin a sin ß sin a sin B = 1+ tan atan B sin a sin ß cos a cos B = 1+ tan atan B -=1+tan atan B -= 1 + tan tan B

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°.
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**Establish the identity.**

\[
\frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = 1 + \tan \alpha \tan \beta
\]

---

Choose the sequence of steps below that verifies the identity.

**A.** 
\[
\frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = \frac{\sin \alpha \sin \beta - \cos \alpha \cos \beta}{\cos \alpha \cos \beta} = \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} - \frac{\cos \alpha \cos \beta}{\cos \alpha \cos \beta} = \tan \alpha \tan \beta - 1 \neq 1 + \tan \alpha \tan \beta
\]

**B.** 
\[
\frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 - \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 - \tan \alpha \tan \beta \neq 1 + \tan \alpha \tan \beta
\]

**C.** 
\[
\frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 + \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 + \tan \alpha \tan \beta
\]

**D.** 
\[
\frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\cos \alpha \cos \beta} = \frac{\cos \alpha \cos \beta}{\cos \alpha \cos \beta} + \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 + \tan \alpha \tan \beta
\]
Transcribed Image Text:**Establish the identity.** \[ \frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = 1 + \tan \alpha \tan \beta \] --- Choose the sequence of steps below that verifies the identity. **A.** \[ \frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = \frac{\sin \alpha \sin \beta - \cos \alpha \cos \beta}{\cos \alpha \cos \beta} = \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} - \frac{\cos \alpha \cos \beta}{\cos \alpha \cos \beta} = \tan \alpha \tan \beta - 1 \neq 1 + \tan \alpha \tan \beta \] **B.** \[ \frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 - \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 - \tan \alpha \tan \beta \neq 1 + \tan \alpha \tan \beta \] **C.** \[ \frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 + \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 + \tan \alpha \tan \beta \] **D.** \[ \frac{\cos(\alpha - \beta)}{\cos \alpha \cos \beta} = \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\cos \alpha \cos \beta} = \frac{\cos \alpha \cos \beta}{\cos \alpha \cos \beta} + \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 1 + \tan \alpha \tan \beta \]
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