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 find the value of the expression \( \cot^{-1}(\alpha) + \cot^{-1}\left(\frac{1}{\alpha}\right) - \frac{\pi}{2} \) given that \( \alpha \) is a real root of the equation \( x^2 + 3x - 10 = 0 \).

### Step 1: Solve the quadratic equation

The given quadratic equation is:

\[ x^2 + 3x - 10 = 0 \]

We can use the quadratic formula to find the roots:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = 3 \), and \( c = -10 \).

### Step 2: Calculate the discriminant

First, we calculate the discriminant:

\[ b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49 \]

### Step 3: Find the roots

Now, substituting back into the quadratic formula:

\[ x = \frac{-3 \pm \sqrt{49}}{2 \cdot 1} = \frac{-3 \pm 7}{2} \]

Calculating the two possible values:

1. \( x_1 = \frac{-3 + 7}{2} = \frac{4}{2} = 2 \)
2. \( x_2 = \frac{-3 - 7}{2} = \frac{-10}{2} = -5 \)

Thus, the roots are \( \alpha = 2 \) and \( \beta = -5 \).

### Step 4: Evaluate the expression

Now we need to evaluate:

\[ \cot^{-1}(\alpha) + \cot^{-1}\left(\frac{1}{\alpha}\right) - \frac{\pi}{2} \]

Substituting \( \alpha = 2 \):

\[ \cot^{-1}(2) + \cot^{-1}\left(\frac{1}{2}\right) - \frac{\pi}{2} \]

Using the identity \( \cot^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \):

\[ \cot^{-1}(2) + \cot^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2} \]

Thus, we have:

\[ \frac{\pi}{2} - \frac{\pi}{2} = 0 \]

### Final Answer

Therefore, the value of the expression is:

\[ \boxed{0} \]