Prove: ` 4 tan^(-1) (1/5 )- tan^(-1)( 1/239) = pi/4`

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(2)`

Answer

To prove the equation \( 4 \tan^{-1} \left( \frac{1}{5} \right) - \tan^{-1} \left( \frac{1}{239} \right) = \frac{\pi}{4} \), we will follow a systematic approach using trigonometric identities.

### Step-by-Step Solution:

1. **Start with the Left-Hand Side (LHS)**:
\[ LHS = 4 \tan^{-1} \left( \frac{1}{5} \right) - \tan^{-1} \left( \frac{1}{239} \right) \]

2. **Use the Double Angle Formula**:
We can rewrite \( 4 \tan^{-1} \left( \frac{1}{5} \right) \) as \( 2 \times 2 \tan^{-1} \left( \frac{1}{5} \right) \). Using the formula \( 2 \tan^{-1}(x) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \), we first calculate \( 2 \tan^{-1} \left( \frac{1}{5} \right) \):
\[ 2 \tan^{-1} \left( \frac{1}{5} \right) = \tan^{-1} \left( \frac{2 \cdot \frac{1}{5}}{1 - \left( \frac{1}{5} \right)^2} \right) \]
Simplifying this, we have:
\[ = \tan^{-1} \left( \frac{\frac{2}{5}}{1 - \frac{1}{25}} \right) = \tan^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} \right) = \tan^{-1} \left( \frac{2 \cdot 25}{5 \cdot 24} \right) = \tan^{-1} \left( \frac{10}{24} \right) = \tan^{-1} \left( \frac{5}{12} \right) \]

3. **Calculate \( 4 \tan^{-1} \left( \frac{1}{5} \right) \)**:
Now we apply the double angle formula again:
\[ 4 \tan^{-1} \left( \frac{1}{5} \right) = 2 \cdot 2 \tan^{-1} \left( \frac{1}{5} \right) = 2 \tan^{-1} \left( \frac{5}{12} \right) \]
Using the double angle formula again:
\[ 2 \tan^{-1} \left( \frac{5}{12} \right) = \tan^{-1} \left( \frac{2 \cdot \frac{5}{12}}{1 - \left( \frac{5}{12} \right)^2} \right) \]
Simplifying this:
\[ = \tan^{-1} \left( \frac{\frac{10}{12}}{1 - \frac{25}{144}} \right) = \tan^{-1} \left( \frac{\frac{5}{6}}{\frac{119}{144}} \right) = \tan^{-1} \left( \frac{5 \cdot 144}{6 \cdot 119} \right) = \tan^{-1} \left( \frac{120}{119} \right) \]

4. **Combine with the Second Term**:
Now we substitute back into the LHS:
\[ LHS = \tan^{-1} \left( \frac{120}{119} \right) - \tan^{-1} \left( \frac{1}{239} \right) \]

5. **Use the Formula for the Difference of Arctangents**:
We apply the formula:
\[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1} \left( \frac{A - B}{1 + AB} \right) \]
Here, \( A = \frac{120}{119} \) and \( B = \frac{1}{239} \):
\[ LHS = \tan^{-1} \left( \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \cdot \frac{1}{239}} \right) \]

6. **Simplify the Expression**:
Finding a common denominator for the numerator:
\[ = \tan^{-1} \left( \frac{\frac{120 \cdot 239 - 119}{119 \cdot 239}}{1 + \frac{120}{119 \cdot 239}} \right) \]
The numerator simplifies to:
\[ = \tan^{-1} \left( \frac{28680 - 119}{28441} \right) = \tan^{-1} \left( \frac{28561}{28441} \right) \]

7. **Final Simplification**:
Notice that:
\[ \frac{28561}{28441} = 1 \quad \text{(since \( 28561 = 169^2 \) and \( 28441 = 169^2 - 1 \))} \]
Thus:
\[ LHS = \tan^{-1}(1) = \frac{\pi}{4} \]

8. **Conclusion**:
Therefore, we have shown:
\[ 4 \tan^{-1} \left( \frac{1}{5} \right) - \tan^{-1} \left( \frac{1}{239} \right) = \frac{\pi}{4} \]
Hence, the equation is proved.