If `lambda` is a root of `x^(2) + 3x + 1 = 0` then `tan^(-1) lambda+ tan^(-1) (1/lambda)` is equal to

Options

A. `pi/2`
B. `-pi/2`
C. `pi/3`
D. none of these

Answer

To solve the problem, we need to find the value of \( \tan^{-1}(\lambda) + \tan^{-1}\left(\frac{1}{\lambda}\right) \) where \( \lambda \) is a root of the equation \( x^2 + 3x + 1 = 0 \).

### Step-by-Step Solution:

1. **Find the roots of the quadratic equation**:
The given equation is:
\[ x^2 + 3x + 1 = 0 \]
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 3, c = 1 \). Plugging in these values:
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 4}}{2} = \frac{-3 \pm \sqrt{5}}{2} \]
Thus, the roots are:
\[ \lambda_1 = \frac{-3 + \sqrt{5}}{2}, \quad \lambda_2 = \frac{-3 - \sqrt{5}}{2} \]

2. **Identify the sign of the roots**:
Both roots \( \lambda_1 \) and \( \lambda_2 \) are negative since \( \sqrt{5} < 3 \). Thus, we can denote \( \lambda = \lambda_1 \) or \( \lambda = \lambda_2 \).

3. **Use the identity for the sum of inverse tangents**:
We know that:
\[ \tan^{-1}(\lambda) + \tan^{-1}\left(\frac{1}{\lambda}\right) = \frac{\pi}{2} \quad \text{if } \lambda > 0 \]
However, since \( \lambda \) is negative, we need to modify our approach:
\[ \tan^{-1}(\lambda) + \tan^{-1}\left(\frac{1}{\lambda}\right) = \tan^{-1}(\lambda) + \cot^{-1}(\lambda) \]

4. **Convert cotangent to tangent**:
Since \( \cot^{-1}(\lambda) = \frac{\pi}{2} - \tan^{-1}(\lambda) \):
\[ \tan^{-1}(\lambda) + \cot^{-1}(\lambda) = \tan^{-1}(\lambda) + \left(\frac{\pi}{2} - \tan^{-1}(\lambda)\right) = \frac{\pi}{2} \]

5. **Final result**:
Therefore, we conclude that:
\[ \tan^{-1}(\lambda) + \tan^{-1}\left(\frac{1}{\lambda}\right) = \frac{\pi}{2} \]

### Conclusion:
The value of \( \tan^{-1}(\lambda) + \tan^{-1}\left(\frac{1}{\lambda}\right) \) is:
\[ \boxed{\frac{\pi}{2}} \]