If `S=tan^(-1)(1/3)+tan^(-1)(1/7)+tan^(-1)(1/13)+....` to 10 terms. Find `tanS`

Answer

To solve the problem, we need to find the value of \( S = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) + \ldots \) for 10 terms and then compute \( \tan S \).

### Step-by-Step Solution:

1. **Identify the Series**:
We have the series:
\[ S = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) + \ldots \]
The terms can be expressed as \( \tan^{-1}\left(\frac{1}{n}\right) \) for specific values of \( n \).

2. **Use the Addition Formula for Inverse Tangent**:
Recall the formula:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \quad \text{if } ab < 1 \]
We will apply this formula iteratively to combine the terms.

3. **Pair the Terms**:
Let's pair the terms:
\[ S = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) + \ldots + \tan^{-1}\left(\frac{1}{19}\right) \]
We can express each term as:
\[ \tan^{-1}\left(\frac{1}{n}\right) = \tan^{-1}(a) - \tan^{-1}(b) \]
where \( a \) and \( b \) are chosen such that the argument fits the form of the addition formula.

4. **Write in Terms of Differences**:
We can express:
\[ S = \tan^{-1}(2) - \tan^{-1}(1) + \tan^{-1}(3) - \tan^{-1}(2) + \tan^{-1}(4) - \tan^{-1}(3) + \ldots + \tan^{-1}(11) - \tan^{-1}(10) \]
This leads to a telescoping series.

5. **Simplify the Series**:
After cancellation, we find:
\[ S = \tan^{-1}(11) - \tan^{-1}(1) \]

6. **Combine Using the Addition Formula**:
Now, we can combine these two terms:
\[ S = \tan^{-1}\left(\frac{11 - 1}{1 + 11 \cdot 1}\right) = \tan^{-1}\left(\frac{10}{12}\right) = \tan^{-1}\left(\frac{5}{6}\right) \]

7. **Find \( \tan S \)**:
Thus, we have:
\[ \tan S = \frac{5}{6} \]

### Final Answer:
\[ \tan S = \frac{5}{6} \]