If median AD of a triangle ABC makes angle `(pi)/(6)` with side BC, then the value of `(cot B-cot C)^(2)` is equal to

Options

A. 6
B. 9
C. 12
D. 15

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have a triangle ABC with a median AD. The median AD makes an angle of \(\frac{\pi}{6}\) with the side BC. We need to find the value of \((\cot B - \cot C)^2\).

### Step 2: Apply the MN Theorem
According to the MN theorem, for a triangle with median AD, we have:
\[ M + N \cot \theta = N \cot B - M \cot C \]
where \(M\) and \(N\) are the segments into which the median divides the side BC, and \(\theta\) is the angle between the median and the side BC.

### Step 3: Set \(M = N\)
Since AD is a median, it divides BC into two equal segments. Therefore, we can set \(M = N\). Let \(M = N = m\). The equation simplifies to:
\[ 2m \cot \theta = m \cot B - m \cot C \]
Dividing through by \(m\) (assuming \(m \neq 0\)):
\[ 2 \cot \theta = \cot B - \cot C \]

### Step 4: Substitute \(\theta = \frac{\pi}{6}\)
Now, substitute \(\theta = \frac{\pi}{6}\):
\[ 2 \cot \left(\frac{\pi}{6}\right) = \cot B - \cot C \]
We know that \(\cot \left(\frac{\pi}{6}\right) = \sqrt{3}\), so:
\[ 2 \sqrt{3} = \cot B - \cot C \]

### Step 5: Square the Result
Now we need to find \((\cot B - \cot C)^2\):
\[ (\cot B - \cot C)^2 = (2 \sqrt{3})^2 = 4 \cdot 3 = 12 \]

### Final Answer
Thus, the value of \((\cot B - \cot C)^2\) is \(12\).