Use the information given below to find tan (a+B). 3. sin a = with a in quadrant I 5' 4. with B in quadrant II cos B=- Give the exact answer, not a decimal approximation. tan %3D
Use the information given below to find tan (a+B). 3. sin a = with a in quadrant I 5' 4. with B in quadrant II cos B=- Give the exact answer, not a decimal approximation. tan %3D
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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![### Finding the Exact Value of \( \tan (\alpha + \beta) \)
To solve for \( \tan (\alpha + \beta) \), we're given several key pieces of information:
1. \( \sin \alpha = \frac{3}{5} \), and \(\alpha\) is in quadrant I.
2. \( \cos \beta = -\frac{4}{5} \), and \(\beta\) is in quadrant II.
**Steps:**
1. **Determine additional trigonometric values:**
- For \( \alpha \) in quadrant I:
- \( \sin \alpha = \frac{3}{5} \)
- Using Pythagoras’ theorem: \( \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \)
- For \( \beta \) in quadrant II:
- \( \cos \beta = -\frac{4}{5} \)
- Using Pythagoras’ theorem: \( \sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \left(-\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \)
2. **Calculate \( \tan \alpha \) and \( \tan \beta \):**
- \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \)
- \( \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \)
3. **Use the tangent sum formula:**
\[
\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}
\]
Substituting](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8dbcbf0b-9123-4a1c-9796-89fa34227664%2F231588c3-6d77-4a4f-ac76-6a2c833a607d%2Fgm07nia_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Exact Value of \( \tan (\alpha + \beta) \)
To solve for \( \tan (\alpha + \beta) \), we're given several key pieces of information:
1. \( \sin \alpha = \frac{3}{5} \), and \(\alpha\) is in quadrant I.
2. \( \cos \beta = -\frac{4}{5} \), and \(\beta\) is in quadrant II.
**Steps:**
1. **Determine additional trigonometric values:**
- For \( \alpha \) in quadrant I:
- \( \sin \alpha = \frac{3}{5} \)
- Using Pythagoras’ theorem: \( \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \)
- For \( \beta \) in quadrant II:
- \( \cos \beta = -\frac{4}{5} \)
- Using Pythagoras’ theorem: \( \sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \left(-\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \)
2. **Calculate \( \tan \alpha \) and \( \tan \beta \):**
- \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \)
- \( \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \)
3. **Use the tangent sum formula:**
\[
\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}
\]
Substituting
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