Find the value of `4 tan^-1 (1/5) - tan^-1 (1/239) `

Explanation

To solve the problem \( 4 \tan^{-1} \left( \frac{1}{5} \right) - \tan^{-1} \left( \frac{1}{239} \right) \), we will use the properties and formulas of inverse tangent functions.

### Step-by-Step Solution:

1. **Use the Double Angle Formula for Inverse Tangent**:
The formula for \( 2 \tan^{-1}(x) \) is:
\[ 2 \tan^{-1}(x) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]
Therefore, we can express \( 4 \tan^{-1} \left( \frac{1}{5} \right) \) as:
\[ 4 \tan^{-1} \left( \frac{1}{5} \right) = 2 \cdot 2 \tan^{-1} \left( \frac{1}{5} \right) = 2 \tan^{-1} \left( \frac{2 \cdot \frac{1}{5}}{1 - \left( \frac{1}{5} \right)^2} \right) \]

2. **Calculate \( 2 \tan^{-1} \left( \frac{1}{5} \right) \)**:
First, calculate \( 2 \cdot \frac{1}{5} = \frac{2}{5} \) and \( 1 - \left( \frac{1}{5} \right)^2 = 1 - \frac{1}{25} = \frac{24}{25} \).
Thus,
\[ 2 \tan^{-1} \left( \frac{1}{5} \right) = \tan^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} \right) = \tan^{-1} \left( \frac{2 \cdot 25}{5 \cdot 24} \right) = \tan^{-1} \left( \frac{10}{24} \right) = \tan^{-1} \left( \frac{5}{12} \right) \]

3. **Now apply the formula again**:
We have:
\[ 4 \tan^{-1} \left( \frac{1}{5} \right) = 2 \tan^{-1} \left( \frac{5}{12} \right) \]
Using the double angle formula again:
\[ 2 \tan^{-1} \left( \frac{5}{12} \right) = \tan^{-1} \left( \frac{2 \cdot \frac{5}{12}}{1 - \left( \frac{5}{12} \right)^2} \right) \]

4. **Calculate \( 2 \cdot \frac{5}{12} \) and \( 1 - \left( \frac{5}{12} \right)^2 \)**:
Here, \( 2 \cdot \frac{5}{12} = \frac{10}{12} = \frac{5}{6} \) and \( 1 - \left( \frac{5}{12} \right)^2 = 1 - \frac{25}{144} = \frac{119}{144} \).
Thus,
\[ 2 \tan^{-1} \left( \frac{5}{12} \right) = \tan^{-1} \left( \frac{\frac{5}{6}}{\frac{119}{144}} \right) = \tan^{-1} \left( \frac{5 \cdot 144}{6 \cdot 119} \right) = \tan^{-1} \left( \frac{120}{119} \right) \]

5. **Now we can substitute back into the original expression**:
We have:
\[ 4 \tan^{-1} \left( \frac{1}{5} \right) - \tan^{-1} \left( \frac{1}{239} \right) = \tan^{-1} \left( \frac{120}{119} \right) - \tan^{-1} \left( \frac{1}{239} \right) \]

6. **Use the formula for the difference of two inverse tangents**:
The formula is:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left( \frac{a - b}{1 + ab} \right) \]
Here, \( a = \frac{120}{119} \) and \( b = \frac{1}{239} \).
So,
\[ \tan^{-1} \left( \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \cdot \frac{1}{239}} \right) \]

7. **Calculate the numerator and denominator**:
The numerator:
\[ \frac{120}{119} - \frac{1}{239} = \frac{120 \cdot 239 - 1 \cdot 119}{119 \cdot 239} = \frac{28740 - 119}{28441} = \frac{28721}{28441} \]
The denominator:
\[ 1 + \frac{120}{119} \cdot \frac{1}{239} = 1 + \frac{120}{28441} = \frac{28441 + 120}{28441} = \frac{28561}{28441} \]

8. **Combine the results**:
Thus, we have:
\[ \tan^{-1} \left( \frac{28721}{28561} \right) \]

9. **Recognize that \( \tan^{-1}(1) = \frac{\pi}{4} \)**:
Since \( \frac{28721}{28561} = 1 \), we conclude:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]

### Final Result:
Thus, the value of \( 4 \tan^{-1} \left( \frac{1}{5} \right) - \tan^{-1} \left( \frac{1}{239} \right) \) is:
\[ \frac{\pi}{4} \]