5 12 Find the EXACT value of tan(A + B) if csc A = where A is in Quadrant II and tan B = 3 5 in Quadrant I. Assume all angles are measured from standard position. tan(A + B) = where B is
5 12 Find the EXACT value of tan(A + B) if csc A = where A is in Quadrant II and tan B = 3 5 in Quadrant I. Assume all angles are measured from standard position. tan(A + B) = where B is
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
![**Trigonometry Problem**
*Objective: Determine the exact value of the tangent of the sum of two angles.*
**Problem Statement**
Find the EXACT value of \( \tan(A + B) \) if \(\csc A = \frac{5}{3}\) where \(A\) is in Quadrant II and \(\tan B = \frac{12}{5}\) where \(B\) is in Quadrant I. Assume all angles are measured from standard position.
\[
\tan(A + B) = \text{\_\_\_\_\_\_\_\_}
\]
**Steps to Solve:**
1. **Identify Trigonometric Functions**:
- Given \(\csc A = \frac{5}{3}\), use the reciprocal identity \(\csc A = \frac{1}{\sin A}\) to determine \(\sin A\).
\[
\sin A = \frac{1}{\csc A} = \frac{1}{\frac{5}{3}} = \frac{3}{5}
\]
Since \(A\) is in Quadrant II, \(\sin A\) is positive, and \(\cos A\) is negative. Use the Pythagorean identity to find \(\cos A\):
\[
\cos^2 A + \sin^2 A = 1 \implies \cos^2 A + \left(\frac{3}{5}\right)^2 = 1 \implies \cos^2 A + \frac{9}{25} = 1 \implies \cos^2 A = \frac{16}{25} \implies \cos A = -\frac{4}{5}
\]
- Given \(\tan B = \frac{12}{5}\), use the identity \(\tan B = \frac{\sin B}{\cos B}\) to find \(\sin B\) and \(\cos B\). Since \(B\) is in Quadrant I, both \(\sin B\) and \(\cos B\) are positive.
\[
\tan B = \frac{\sin B}{\cos B} = \frac{12}{5}
\]
Use the Pythagorean identity in tangent form:
\[
1 + \tan^2 B](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b559ab3-5119-482d-94d5-a98a530a62e8%2Fc37dd700-0586-4cba-bd16-ec90e79694c1%2Fx1zp8rg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Trigonometry Problem**
*Objective: Determine the exact value of the tangent of the sum of two angles.*
**Problem Statement**
Find the EXACT value of \( \tan(A + B) \) if \(\csc A = \frac{5}{3}\) where \(A\) is in Quadrant II and \(\tan B = \frac{12}{5}\) where \(B\) is in Quadrant I. Assume all angles are measured from standard position.
\[
\tan(A + B) = \text{\_\_\_\_\_\_\_\_}
\]
**Steps to Solve:**
1. **Identify Trigonometric Functions**:
- Given \(\csc A = \frac{5}{3}\), use the reciprocal identity \(\csc A = \frac{1}{\sin A}\) to determine \(\sin A\).
\[
\sin A = \frac{1}{\csc A} = \frac{1}{\frac{5}{3}} = \frac{3}{5}
\]
Since \(A\) is in Quadrant II, \(\sin A\) is positive, and \(\cos A\) is negative. Use the Pythagorean identity to find \(\cos A\):
\[
\cos^2 A + \sin^2 A = 1 \implies \cos^2 A + \left(\frac{3}{5}\right)^2 = 1 \implies \cos^2 A + \frac{9}{25} = 1 \implies \cos^2 A = \frac{16}{25} \implies \cos A = -\frac{4}{5}
\]
- Given \(\tan B = \frac{12}{5}\), use the identity \(\tan B = \frac{\sin B}{\cos B}\) to find \(\sin B\) and \(\cos B\). Since \(B\) is in Quadrant I, both \(\sin B\) and \(\cos B\) are positive.
\[
\tan B = \frac{\sin B}{\cos B} = \frac{12}{5}
\]
Use the Pythagorean identity in tangent form:
\[
1 + \tan^2 B
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