Given: cos A = 7/25,           270° < A < 360°
sin B = -12/13,         180° < B < 270°

1. tan (A + B)

2. tan (A - B)

Given the trigonometric identities and the information:

\[
\cos A = \frac{7}{25}, \quad 270^\circ < A < 360^\circ \quad (\text{fourth quadrant})
\]
\[
\sin B = -\frac{12}{13}, \quad 180^\circ < B < 270^\circ \quad (\text{third quadrant})
\]

We can calculate \( \tan(A + B) \) and \( \tan(A - B) \) using the following identities for tangent sum and difference:

\[
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}
\]
\[
\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}
\]

### Step 1: Calculate \( \sin A \) and \( \cos B \)

#### For angle \( A \) (fourth quadrant):
We know that \( \sin^2 A + \cos^2 A = 1 \), so:
\[
\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{7}{25}\right)^2 = 1 - \frac{49}{625} = \frac{576}{625}
\]
\[
\sin A = -\frac{24}{25} \quad (\text{since } A \text{ is in the fourth quadrant, where sine is negative})
\]

#### For angle \( B \) (third quadrant):
We know that \( \sin^2 B + \cos^2 B = 1 \), so:
\[
\cos^2 B = 1 - \sin^2 B = 1 - \left(\frac{-12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}
\]
\[
\cos B = -\frac{5}{13} \quad (\text{since } B \text{ is in the third quadrant, where cosine is negative})
\]

### Step 2: Calculate \( \tan A \) and \( \tan B \)

\[
\tan A = \frac{\sin A}{\cos A} = \frac{-\frac{24}{25}}{\frac{7}{25}} = -\frac{24}{7}
\]
\[
\tan B = \frac{\sin B}{\cos B} = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{5}
\]

### Step 3: Calculate \( \tan(A + B) \)

\[
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} = \frac{-\frac{24}{7} + \frac{12}{5}}{1 - \left(-\frac{24}{7}\right) \cdot \frac{12}{5}}
\]

To simplify:

\[
\tan(A + B) = \frac{-\frac{120}{35} + \frac{84}{35}}{1 - \left(-\frac{288}{35}\right)} = \frac{-\frac{36}{35}}{1 + \frac{288}{35}}
\]
\[
\tan(A + B) = \frac{-\frac{36}{35}}{\frac{323}{35}} = \frac{-36}{323}
\]

Thus:
\[
\tan(A + B) = \frac{-36}{323}
\]

### Step 4: Calculate \( \tan(A - B) \)

\[
\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B} = \frac{-\frac{24}{7} - \frac{12}{5}}{1 + \left(-\frac{24}{7}\right) \cdot \frac{12}{5}}
\]

To simplify:

\[
\tan(A - B) = \frac{-\frac{120}{35} - \frac{84}{35}}{1 + \left(-\frac{288}{35}\right)} = \frac{-\frac{204}{35}}{1 - \frac{288}{35}} = \frac{-\frac{204}{35}}{\frac{-253}{35}}
\]
\[
\tan(A - B) = \frac{-204}{-253} = \frac{204}{253}
\]

Thus:
\[
\tan(A - B) = \frac{204}{253}
\]

### Final answers:
1. \( \tan(A + B) = \frac{-36}{323} \)
2. \( \tan(A - B) = \frac{204}{253} \)

---

Do you have any questions or need further clarifications?

Here are 5 related questions:
1. How do you determine the signs of trigonometric functions in different quadrants?
2. Can you use other methods to calculate trigonometric identities for sum and difference of angles?
3. What are the common properties of the unit circle that help in solving trigonometric problems?
4. How does the Pythagorean identity help in finding unknown trigonometric values?
5. What are the practical applications of tangent sum and difference identities?

**Tip:** Always double-check the quadrant to ensure the correct sign for trigonometric functions like sine and cosine.

Given: cos A = 7/25, 270° < A < 360°, sin B = -12/13, 180° < B < 270°. Find: 1. tan(A + B) 2. tan(A - B)

This problem explores how to find tan(A + B) and tan(A - B) given specific values of cos A and sin B in different quadrants. The solution covers key trigonometric identities and Pythagorean identities, suitable for students in Grades 10-12.

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