


     **Given: sin A = 15/17,         0° < A < 90°**

                  **cos B = -5/13,        90° < B < 180°**




**1. tan (A + B)**

**2. tan (A - B)**

Given the following:

- \( \sin A = \frac{15}{17} \), where \( 0^\circ < A < 90^\circ \).
- \( \cos B = -\frac{5}{13} \), where \( 90^\circ < B < 180^\circ \).

We need to calculate:
1. \( \tan(A + B) \)
2. \( \tan(A - B) \)

### Step 1: Find \( \cos A \) and \( \sin B \)
1. **For angle \( A \):**
   - \( \sin A = \frac{15}{17} \)
   - Using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \), we can calculate \( \cos A \).

\[
\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{15}{17}\right)^2} = \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17}
\]

2. **For angle \( B \):**
   - \( \cos B = -\frac{5}{13} \)
   - Using the Pythagorean identity \( \sin^2 B + \cos^2 B = 1 \), we can calculate \( \sin B \).

\[
\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{-5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}
\]

(Note that \( \sin B \) is positive since \( B \) lies in the second quadrant.)

### Step 2: Find \( \tan A \) and \( \tan B \)
1. **For angle \( A \):**
   \[
   \tan A = \frac{\sin A}{\cos A} = \frac{\frac{15}{17}}{\frac{8}{17}} = \frac{15}{8}
   \]

2. **For angle \( B \):**
   \[
   \tan B = \frac{\sin B}{\cos B} = \frac{\frac{12}{13}}{\frac{-5}{13}} = -\frac{12}{5}
   \]

### Step 3: Use the tangent addition and subtraction formulas
The formulas for \( \tan(A + B) \) and \( \tan(A - B) \) are:
\[
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}
\]
\[
\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}
\]

### 1. Calculate \( \tan(A + B) \)
\[
\tan(A + B) = \frac{\frac{15}{8} + \left(-\frac{12}{5}\right)}{1 - \frac{15}{8} \cdot \left(-\frac{12}{5}\right)}
\]
First, simplify the numerator and the denominator:
- Numerator: \( \frac{15}{8} - \frac{12}{5} = \frac{15 \cdot 5 - 12 \cdot 8}{8 \cdot 5} = \frac{75 - 96}{40} = \frac{-21}{40} \)
- Denominator: \( 1 - \frac{15}{8} \cdot \left(-\frac{12}{5}\right) = 1 + \frac{15 \cdot 12}{8 \cdot 5} = 1 + \frac{180}{40} = 1 + 4.5 = 5.5 \)

Thus:
\[
\tan(A + B) = \frac{\frac{-21}{40}}{5.5} = \frac{-21}{40 \times 5.5} = \frac{-21}{220}
\]
So,
\[
\tan(A + B) = \frac{-21}{220}
\]

### 2. Calculate \( \tan(A - B) \)
\[
\tan(A - B) = \frac{\frac{15}{8} - \left(-\frac{12}{5}\right)}{1 + \frac{15}{8} \cdot \left(-\frac{12}{5}\right)}
\]
First, simplify the numerator and the denominator:
- Numerator: \( \frac{15}{8} + \frac{12}{5} = \frac{15 \cdot 5 + 12 \cdot 8}{8 \cdot 5} = \frac{75 + 96}{40} = \frac{171}{40} \)
- Denominator: \( 1 + \frac{15}{8} \cdot \left(-\frac{12}{5}\right) = 1 - \frac{180}{40} = 1 - 4.5 = -3.5 \)

Thus:
\[
\tan(A - B) = \frac{\frac{171}{40}}{-3.5} = \frac{171}{40 \times -3.5} = \frac{171}{-140}
\]
So,
\[
\tan(A - B) = \frac{-171}{140}
\]

### Final Answers:
1. \( \tan(A + B) = \frac{-21}{220} \)
2. \( \tan(A - B) = \frac{-171}{140} \)

Would you like further details or have any other questions? Here are some follow-up questions:

1. What is the value of \( \tan(A + B) \) in decimal form?
2. How does the quadrant of angle \( B \) affect the sign of trigonometric functions?
3. Can you derive the secant of angle \( A \)?
4. What are the general properties of the tangent addition and subtraction formulas?
5. How do you find the exact angle values for \( A \) and \( B \) from the given trigonometric ratios?

**Tip:** Understanding the signs of trigonometric functions in different quadrants is crucial for solving such problems correctly.

Given: sin A = 15/17, 0° < A < 90°, cos B = -5/13, 90° < B < 180°. 1. tan(A + B), 2. tan(A - B)

Learn how to calculate tan(A + B) and tan(A - B) given sin A = 15/17 (0° < A < 90°) and cos B = -5/13 (90° < B < 180°). This problem covers advanced trigonometry concepts, including the use of the tangent addition and subtraction formulas, and the Pythagorean identity. Suitable for Grades 11-12.

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