If Tan `A=15/8` and Tan `B=7/24`, then Cos `(A+B)=?`

Options

A. `87/(425)`
B. `304/425`
C. `297/425`
D. `416/425`

Answer

To solve the problem, we need to find \( \cos(A + B) \) given that \( \tan A = \frac{15}{8} \) and \( \tan B = \frac{7}{24} \).

### Step-by-Step Solution:

1. **Use the Cosine Addition Formula**:
The formula for \( \cos(A + B) \) is:
\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]

2. **Find \( \sin A \) and \( \cos A \)**:
Given \( \tan A = \frac{15}{8} \), we can represent this in a right triangle where:
- Opposite side (perpendicular) = 15
- Adjacent side (base) = 8

To find the hypotenuse \( h \):
\[ h = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \]

Now, we can find \( \sin A \) and \( \cos A \):
\[ \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{15}{17} \]
\[ \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{8}{17} \]

3. **Find \( \sin B \) and \( \cos B \)**:
Given \( \tan B = \frac{7}{24} \), we can represent this in a right triangle where:
- Opposite side (perpendicular) = 7
- Adjacent side (base) = 24

To find the hypotenuse \( h \):
\[ h = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]

Now, we can find \( \sin B \) and \( \cos B \):
\[ \sin B = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{7}{25} \]
\[ \cos B = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{24}{25} \]

4. **Substitute Values into the Cosine Addition Formula**:
Now we substitute \( \sin A \), \( \cos A \), \( \sin B \), and \( \cos B \) into the formula:
\[ \cos(A + B) = \left(\frac{8}{17} \cdot \frac{24}{25}\right) - \left(\frac{15}{17} \cdot \frac{7}{25}\right) \]

5. **Calculate Each Term**:
- First term:
\[ \frac{8 \cdot 24}{17 \cdot 25} = \frac{192}{425} \]
- Second term:
\[ \frac{15 \cdot 7}{17 \cdot 25} = \frac{105}{425} \]

6. **Combine the Results**:
\[ \cos(A + B) = \frac{192}{425} - \frac{105}{425} = \frac{192 - 105}{425} = \frac{87}{425} \]

### Final Answer:
\[ \cos(A + B) = \frac{87}{425} \]