If `alpha` is a real root of the equation `x^(2)+3x-tan2=0` then `cot^(-1)alpha+"cot"^(-1)1/(alpha)-(pi)/2` can be equal to

Options

A. `0`
B. `(pi)/2`
C. `pi`
D. `(3pi)/2`

Explanation

To solve the problem, we need to evaluate the expression \( \cot^{-1} \alpha + \cot^{-1} \frac{1}{\alpha} - \frac{\pi}{2} \), given that \( \alpha \) is a real root of the equation \( x^2 + 3x - 10 = 0 \).

### Step-by-Step Solution:

1. **Find the roots of the equation**:
We start with the quadratic equation:
\[ x^2 + 3x - 10 = 0 \]
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 3, c = -10 \).

\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \]
\[ = \frac{-3 \pm \sqrt{9 + 40}}{2} \]
\[ = \frac{-3 \pm \sqrt{49}}{2} \]
\[ = \frac{-3 \pm 7}{2} \]

This gives us two roots:
\[ x_1 = \frac{4}{2} = 2, \quad x_2 = \frac{-10}{2} = -5 \]

Since we are interested in the real root, we can take \( \alpha = 2 \) (as it is a positive real root).

2. **Evaluate \( \cot^{-1} \alpha + \cot^{-1} \frac{1}{\alpha} \)**:
We know that:
\[ \cot^{-1} \alpha + \cot^{-1} \frac{1}{\alpha} = \frac{\pi}{2} \]
This is a standard result from inverse trigonometric identities.

3. **Substituting into the expression**:
Now, we substitute this result into our expression:
\[ \cot^{-1} \alpha + \cot^{-1} \frac{1}{\alpha} - \frac{\pi}{2} = \frac{\pi}{2} - \frac{\pi}{2} = 0 \]

4. **Conclusion**:
Therefore, the value of \( \cot^{-1} \alpha + \cot^{-1} \frac{1}{\alpha} - \frac{\pi}{2} \) is:
\[ \boxed{0} \]