Trigonometry (11th Edition)
11th Edition
ISBN: 9780134217437
Author: Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher: PEARSON
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![### Trigonometry Problems and Identities
#### 4. Add the two fractions and simplify the result:
\[ \frac{\cos x}{\sec x} + \frac{\sin x}{\csc x} \]
**Solution:**
To simplify the given expression, we follow these steps:
1. Recall that \(\sec x\) is the reciprocal of \(\cos x\) and \(\csc x\) is the reciprocal of \(\sin x\). Therefore:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \csc x = \frac{1}{\sin x} \]
2. Substitute these values into the fractions:
\[ \frac{\cos x}{\sec x} = \frac{\cos x}{\frac{1}{\cos x}} = \cos^2 x \]
\[ \frac{\sin x}{\csc x} = \frac{\sin x}{\frac{1}{\sin x}} = \sin^2 x \]
3. Add the two simplified fractions:
\[ \cos^2 x + \sin^2 x \]
4. Recall the Pythagorean identity for sine and cosine:
\[ \cos^2 x + \sin^2 x = 1 \]
Therefore, the simplified expression is:
\[ \boxed{1} \]
#### 5. Verify that the expression is an identity:
\[ \frac{\sin^2 \theta}{\cos \theta} = \sec \theta - \cos \theta \]
**Verification:**
To verify this identity, we perform the following steps:
1. Express \(\sec \theta\) in terms of \(\cos \theta\):
\[ \sec \theta = \frac{1}{\cos \theta} \]
2. Substitute this value into the right side of the equation:
\[ \sec \theta - \cos \theta = \frac{1}{\cos \theta} - \cos \theta \]
3. Combine the terms over a common denominator:
\[ \frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta} \]
4. Use the Pythagorean identity \(\sin^2 \theta](https://content.bartleby.com/qna-images/question/681a1dd5-ea79-4df0-af09-f7e014c18f3f/824377e6-8373-4eef-89d0-0c01f1965366/azpfwvj_processed.jpeg)
Transcribed Image Text:### Trigonometry Problems and Identities
#### 4. Add the two fractions and simplify the result:
\[ \frac{\cos x}{\sec x} + \frac{\sin x}{\csc x} \]
**Solution:**
To simplify the given expression, we follow these steps:
1. Recall that \(\sec x\) is the reciprocal of \(\cos x\) and \(\csc x\) is the reciprocal of \(\sin x\). Therefore:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \csc x = \frac{1}{\sin x} \]
2. Substitute these values into the fractions:
\[ \frac{\cos x}{\sec x} = \frac{\cos x}{\frac{1}{\cos x}} = \cos^2 x \]
\[ \frac{\sin x}{\csc x} = \frac{\sin x}{\frac{1}{\sin x}} = \sin^2 x \]
3. Add the two simplified fractions:
\[ \cos^2 x + \sin^2 x \]
4. Recall the Pythagorean identity for sine and cosine:
\[ \cos^2 x + \sin^2 x = 1 \]
Therefore, the simplified expression is:
\[ \boxed{1} \]
#### 5. Verify that the expression is an identity:
\[ \frac{\sin^2 \theta}{\cos \theta} = \sec \theta - \cos \theta \]
**Verification:**
To verify this identity, we perform the following steps:
1. Express \(\sec \theta\) in terms of \(\cos \theta\):
\[ \sec \theta = \frac{1}{\cos \theta} \]
2. Substitute this value into the right side of the equation:
\[ \sec \theta - \cos \theta = \frac{1}{\cos \theta} - \cos \theta \]
3. Combine the terms over a common denominator:
\[ \frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta} \]
4. Use the Pythagorean identity \(\sin^2 \theta
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