#11 Find tan (A - B). 3 cos A = 1 sinB = < A< T 2 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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can you help me solve these 2 questions
![**Question 11: Find \(\tan(A - B)\).**
Given:
- \(\cos A = -\frac{1}{5}\)
- \(\sin B = \frac{3}{5}\)
Constraints:
- \(\frac{\pi}{2} < A < \pi\)
- \(\frac{\pi}{2} < B < \pi\)
**Solution Approach:**
To find \(\tan(A - B)\), we can use the formula for the tangent of a difference:
\[
\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}
\]
First, we need to find \(\sin A\) and \(\cos B\) using the identities:
\[
\sin^2 A + \cos^2 A = 1
\]
\[
\sin^2 B + \cos^2 B = 1
\]
Given \(\cos A = -\frac{1}{5}\):
\[
\sin^2 A = 1 - \left(-\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25}
\]
Since \(\frac{\pi}{2} < A < \pi\), \(\sin A\) is positive:
\[
\sin A = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}
\]
Now, for \(\cos B\):
\[
\cos^2 B = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}
\]
Since \(\frac{\pi}{2} < B < \pi\), \(\cos B\) is negative:
\[
\cos B = -\sqrt{\frac{16}{25}} = -\frac{4}{5}
\]
Now, we can find \(\tan A\) and \(\tan B\):
\[
\tan A = \frac{\sin A}{\cos A} = \frac{\frac{2\sqrt{6}}{5}}{-\frac{1}{5}} = -2\sqrt{6}
\]
\[
\tan B = \frac{\sin B}{\cos](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ed323b9-8e35-4824-8e78-fcaf55c6ad57%2F98beb658-053f-4abe-877d-3cd0a09e005c%2Fu5dp9qc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 11: Find \(\tan(A - B)\).**
Given:
- \(\cos A = -\frac{1}{5}\)
- \(\sin B = \frac{3}{5}\)
Constraints:
- \(\frac{\pi}{2} < A < \pi\)
- \(\frac{\pi}{2} < B < \pi\)
**Solution Approach:**
To find \(\tan(A - B)\), we can use the formula for the tangent of a difference:
\[
\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}
\]
First, we need to find \(\sin A\) and \(\cos B\) using the identities:
\[
\sin^2 A + \cos^2 A = 1
\]
\[
\sin^2 B + \cos^2 B = 1
\]
Given \(\cos A = -\frac{1}{5}\):
\[
\sin^2 A = 1 - \left(-\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25}
\]
Since \(\frac{\pi}{2} < A < \pi\), \(\sin A\) is positive:
\[
\sin A = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}
\]
Now, for \(\cos B\):
\[
\cos^2 B = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}
\]
Since \(\frac{\pi}{2} < B < \pi\), \(\cos B\) is negative:
\[
\cos B = -\sqrt{\frac{16}{25}} = -\frac{4}{5}
\]
Now, we can find \(\tan A\) and \(\tan B\):
\[
\tan A = \frac{\sin A}{\cos A} = \frac{\frac{2\sqrt{6}}{5}}{-\frac{1}{5}} = -2\sqrt{6}
\]
\[
\tan B = \frac{\sin B}{\cos
![**Problem #12**
**Objective:** Find \(\tan(A + B)\).
**Given:**
- \(\cos A = \frac{15}{17}\)
- \(\sin B = \frac{4}{5}\)
**Constraints:**
- \(\frac{\pi}{2} < A < \pi\)
- \(0 < B < \frac{\pi}{2}\)
**Explanation:**
To solve this, you'll need to use the formula for \(\tan(A + B)\):
\[
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
\]
1. **Finding \(\tan A\):**
Since \(\cos A = \frac{15}{17}\) and \(A\) is in the second quadrant (\(\frac{\pi}{2} < A < \pi\)), \(\sin A\) will be positive but less than \(1\). Use the identity \(\sin^2 A + \cos^2 A = 1\):
\[
\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{15}{17}\right)^2} = \sqrt{\frac{64}{289}} = \frac{8}{17}
\]
\(\tan A = \frac{\sin A}{\cos A} = \frac{8}{15}\)
2. **Finding \(\tan B\):**
Since \(B\) is in the first quadrant, \(\tan B = \frac{\sin B}{\cos B}\). Use the identity \(\sin^2 B + \cos^2 B = 1\):
\[
\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25}} = \frac{3}{5}
\]
\(\tan B = \frac{\sin B}{\cos B} = \frac{4}{3}\)
3. **Calculate \(\tan(A + B)\):**
\[
\tan(A + B) = \frac{\frac{8}{15} + \frac{4}{3}}{1 - \frac{8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ed323b9-8e35-4824-8e78-fcaf55c6ad57%2F98beb658-053f-4abe-877d-3cd0a09e005c%2Ftal6t8p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem #12**
**Objective:** Find \(\tan(A + B)\).
**Given:**
- \(\cos A = \frac{15}{17}\)
- \(\sin B = \frac{4}{5}\)
**Constraints:**
- \(\frac{\pi}{2} < A < \pi\)
- \(0 < B < \frac{\pi}{2}\)
**Explanation:**
To solve this, you'll need to use the formula for \(\tan(A + B)\):
\[
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
\]
1. **Finding \(\tan A\):**
Since \(\cos A = \frac{15}{17}\) and \(A\) is in the second quadrant (\(\frac{\pi}{2} < A < \pi\)), \(\sin A\) will be positive but less than \(1\). Use the identity \(\sin^2 A + \cos^2 A = 1\):
\[
\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{15}{17}\right)^2} = \sqrt{\frac{64}{289}} = \frac{8}{17}
\]
\(\tan A = \frac{\sin A}{\cos A} = \frac{8}{15}\)
2. **Finding \(\tan B\):**
Since \(B\) is in the first quadrant, \(\tan B = \frac{\sin B}{\cos B}\). Use the identity \(\sin^2 B + \cos^2 B = 1\):
\[
\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25}} = \frac{3}{5}
\]
\(\tan B = \frac{\sin B}{\cos B} = \frac{4}{3}\)
3. **Calculate \(\tan(A + B)\):**
\[
\tan(A + B) = \frac{\frac{8}{15} + \frac{4}{3}}{1 - \frac{8
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