If `theta = cot^(-1) 7+cot^(-1) 8+cot^(-1) 18, " then " cot theta` is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to find the value of \( \cot \theta \) given that \( \theta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18) \).

### Step-by-Step Solution:

1. **Convert \( \cot^{-1} \) to \( \tan^{-1} \)**:
We know that:
\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \]
Therefore, we can rewrite \( \theta \) as:
\[ \theta = \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{8}\right) + \tan^{-1}\left(\frac{1}{18}\right) \]

2. **Use the formula for the sum of arctangents**:
The formula for the sum of two arctangents is:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \quad \text{if } xy < 1 \]
First, we will combine the first two terms:
\[ \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{8}\right) = \tan^{-1}\left(\frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \cdot \frac{1}{8}}\right) \]

3. **Calculate the numerator and denominator**:
- **Numerator**:
\[ \frac{1}{7} + \frac{1}{8} = \frac{8 + 7}{56} = \frac{15}{56} \]
- **Denominator**:
\[ 1 - \frac{1}{7} \cdot \frac{1}{8} = 1 - \frac{1}{56} = \frac{55}{56} \]

4. **Combine the results**:
Thus,
\[ \tan^{-1}\left(\frac{\frac{15}{56}}{\frac{55}{56}}\right) = \tan^{-1}\left(\frac{15}{55}\right) = \tan^{-1}\left(\frac{3}{11}\right) \]

5. **Now add the third term**:
Now we need to add \( \tan^{-1}\left(\frac{3}{11}\right) + \tan^{-1}\left(\frac{1}{18}\right) \):
\[ \tan^{-1}\left(\frac{3}{11}\right) + \tan^{-1}\left(\frac{1}{18}\right) = \tan^{-1}\left(\frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11} \cdot \frac{1}{18}}\right) \]

6. **Calculate the new numerator and denominator**:
- **Numerator**:
\[ \frac{3}{11} + \frac{1}{18} = \frac{3 \cdot 18 + 1 \cdot 11}{198} = \frac{54 + 11}{198} = \frac{65}{198} \]
- **Denominator**:
\[ 1 - \frac{3}{11} \cdot \frac{1}{18} = 1 - \frac{3}{198} = \frac{195}{198} \]

7. **Combine the results**:
Thus,
\[ \tan^{-1}\left(\frac{\frac{65}{198}}{\frac{195}{198}}\right) = \tan^{-1}\left(\frac{65}{195}\right) = \tan^{-1}\left(\frac{13}{39}\right) = \tan^{-1}\left(\frac{1}{3}\right) \]

8. **Find \( \cot \theta \)**:
Since \( \theta = \tan^{-1}\left(\frac{1}{3}\right) \), we have:
\[ \cot \theta = \frac{1}{\tan \theta} = 3 \]

### Final Answer:
\[ \cot \theta = 3 \]