In triangle ABC, if `|[1,1,1], [cot (A/2), cot(B/2), cot(C/2)], [tan(B/2)+tan(C/2), tan(C/2)+ tan(A/2), tan(A/2)+ tan(B/2)]|` then the triangle must be (A) Equilateral (B) Isoceless (C) Right Angle (D) none of these

Options

A. equilateral
B. isosceles
C. obtuse angled
D. none of these

Explanation

To solve the given problem, we need to evaluate the determinant of the matrix formed by the given expressions and analyze the conditions under which the determinant equals zero. The determinant is given by:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ \cot \left(\frac{A}{2}\right) & \cot \left(\frac{B}{2}\right) & \cot \left(\frac{C}{2}\right) \\ \tan \left(\frac{B}{2}\right) + \tan \left(\frac{C}{2}\right) & \tan \left(\frac{C}{2}\right) + \tan \left(\frac{A}{2}\right) & \tan \left(\frac{A}{2}\right) + \tan \left(\frac{B}{2}\right) \end{vmatrix} \]

### Step 1: Perform Column Operations
We will perform some column operations to simplify the determinant. Let's modify the first two columns:

1. **C1 = C1 - C2**
2. **C2 = C2 - C3**

After performing these operations, we get:

\[ D = \begin{vmatrix} 1 & 0 & 1 \\ \cot \left(\frac{A}{2}\right) - \cot \left(\frac{B}{2}\right) & \cot \left(\frac{B}{2}\right) - \cot \left(\frac{C}{2}\right) & \cot \left(\frac{C}{2}\right) \\ \tan \left(\frac{B}{2}\right) - \tan \left(\frac{A}{2}\right) & \tan \left(\frac{C}{2}\right) - \tan \left(\frac{B}{2}\right) & \tan \left(\frac{A}{2}\right) + \tan \left(\frac{B}{2}\right) \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Next, we can factor out common terms from the first two columns. We can factor out \((\cot \left(\frac{A}{2}\right) - \cot \left(\frac{B}{2}\right))\) and \((\cot \left(\frac{B}{2}\right) - \cot \left(\frac{C}{2}\right))\):

\[ D = (\cot \left(\frac{A}{2}\right) - \cot \left(\frac{B}{2}\right))(\cot \left(\frac{B}{2}\right) - \cot \left(\frac{C}{2}\right)) \cdot \text{determinant} \]

### Step 3: Evaluate the Determinant
Now we evaluate the determinant of the remaining matrix:

\[ \begin{vmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ \tan \left(\frac{B}{2}\right) - \tan \left(\frac{A}{2}\right) & \tan \left(\frac{C}{2}\right) - \tan \left(\frac{B}{2}\right) & \tan \left(\frac{A}{2}\right) + \tan \left(\frac{B}{2}\right) \end{vmatrix} \]

### Step 4: Analyze the Conditions
For the determinant \(D\) to equal zero, either:

1. \(\cot \left(\frac{A}{2}\right) = \cot \left(\frac{B}{2}\right)\) which implies \(A = B\)
2. \(\cot \left(\frac{B}{2}\right) = \cot \left(\frac{C}{2}\right)\) which implies \(B = C\)
3. \(\tan \left(\frac{B}{2}\right) - \tan \left(\frac{A}{2}\right) = 0\) which implies \(A = B\)

From these conditions, we conclude that at least two angles of the triangle are equal, which means triangle ABC is isosceles.

### Final Answer
Thus, the triangle must be **Isosceles** (Option B).

---