If cos `A =(21)/(29)` and A lies in the fourth quadrant , find `"sin" A and tan A`.
If `"sin" A=(21)/(29)` and A lies in the second quadrant prove that sec `A + tan A=(-5)/(2)`

Answer

To solve the given problems step by step, we will break down each part of the question.

### Part 1: Given \( \cos A = \frac{21}{29} \) and \( A \) lies in the fourth quadrant, find \( \sin A \) and \( \tan A \).

1. **Identify the known values**:
- \( \cos A = \frac{21}{29} \)
- In the fourth quadrant, \( \sin A \) is negative.

2. **Use the Pythagorean identity**:
\[ \sin^2 A + \cos^2 A = 1 \]
Substitute \( \cos A \):
\[ \sin^2 A + \left(\frac{21}{29}\right)^2 = 1 \]

3. **Calculate \( \cos^2 A \)**:
\[ \cos^2 A = \left(\frac{21}{29}\right)^2 = \frac{441}{841} \]

4. **Substitute and solve for \( \sin^2 A \)**:
\[ \sin^2 A + \frac{441}{841} = 1 \]
\[ \sin^2 A = 1 - \frac{441}{841} = \frac{841 - 441}{841} = \frac{400}{841} \]

5. **Take the square root to find \( \sin A \)**:
\[ \sin A = -\sqrt{\frac{400}{841}} = -\frac{20}{29} \]
(Negative because \( A \) is in the fourth quadrant)

6. **Find \( \tan A \)**:
\[ \tan A = \frac{\sin A}{\cos A} = \frac{-\frac{20}{29}}{\frac{21}{29}} = -\frac{20}{21} \]

### Summary of Part 1:
- \( \sin A = -\frac{20}{29} \)
- \( \tan A = -\frac{20}{21} \)

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### Part 2: Given \( \sin A = \frac{21}{29} \) and \( A \) lies in the second quadrant, prove that \( \sec A + \tan A = -\frac{5}{2} \).

1. **Identify the known values**:
- \( \sin A = \frac{21}{29} \)
- In the second quadrant, \( \cos A \) is negative.

2. **Use the Pythagorean identity to find \( \cos A \)**:
\[ \sin^2 A + \cos^2 A = 1 \]
Substitute \( \sin A \):
\[ \left(\frac{21}{29}\right)^2 + \cos^2 A = 1 \]

3. **Calculate \( \sin^2 A \)**:
\[ \sin^2 A = \left(\frac{21}{29}\right)^2 = \frac{441}{841} \]

4. **Substitute and solve for \( \cos^2 A \)**:
\[ \frac{441}{841} + \cos^2 A = 1 \]
\[ \cos^2 A = 1 - \frac{441}{841} = \frac{400}{841} \]

5. **Take the square root to find \( \cos A \)**:
\[ \cos A = -\sqrt{\frac{400}{841}} = -\frac{20}{29} \]
(Negative because \( A \) is in the second quadrant)

6. **Find \( \sec A \) and \( \tan A \)**:
\[ \sec A = \frac{1}{\cos A} = \frac{1}{-\frac{20}{29}} = -\frac{29}{20} \]
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{21}{29}}{-\frac{20}{29}} = -\frac{21}{20} \]

7. **Add \( \sec A \) and \( \tan A \)**:
\[ \sec A + \tan A = -\frac{29}{20} - \frac{21}{20} = -\frac{29 + 21}{20} = -\frac{50}{20} = -\frac{5}{2} \]

### Summary of Part 2:
- \( \sec A + \tan A = -\frac{5}{2} \) (Proved)

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