`tan^(-1) . 3/5 +tan^(-1) . 1/4 = pi/2 `

Answer

To solve the equation \( \tan^{-1} \left( \frac{3}{5} \right) + \tan^{-1} \left( \frac{1}{4} \right) = \frac{\pi}{2} \), we can use the formula for the sum of inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]

provided that \( ab < 1 \).

### Step-by-Step Solution:

1. **Identify \( a \) and \( b \)**:
- Let \( a = \frac{3}{5} \) and \( b = \frac{1}{4} \).

2. **Calculate \( ab \)**:
\[ ab = \left( \frac{3}{5} \right) \left( \frac{1}{4} \right) = \frac{3}{20} \]
Since \( \frac{3}{20} < 1 \), we can use the formula.

3. **Apply the formula**:
\[ \tan^{-1} \left( \frac{3}{5} \right) + \tan^{-1} \left( \frac{1}{4} \right) = \tan^{-1} \left( \frac{\frac{3}{5} + \frac{1}{4}}{1 - \frac{3}{5} \cdot \frac{1}{4}} \right) \]

4. **Find the numerator**:
- To add \( \frac{3}{5} \) and \( \frac{1}{4} \), we need a common denominator:
\[ \text{LCM of 5 and 4 is 20.} \]
\[ \frac{3}{5} = \frac{12}{20}, \quad \frac{1}{4} = \frac{5}{20} \]
\[ \text{Numerator: } 12 + 5 = 17 \]

5. **Find the denominator**:
\[ 1 - ab = 1 - \frac{3}{20} = \frac{20 - 3}{20} = \frac{17}{20} \]

6. **Combine the results**:
\[ \tan^{-1} \left( \frac{17/20}{17/20} \right) = \tan^{-1}(1) \]

7. **Evaluate \( \tan^{-1}(1) \)**:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]

8. **Conclusion**:
Since \( \tan^{-1}(1) \) is not equal to \( \frac{\pi}{2} \), the statement is false.

### Final Result:
\[ \tan^{-1} \left( \frac{3}{5} \right) + \tan^{-1} \left( \frac{1}{4} \right) \neq \frac{\pi}{2} \]