In a `Delta ABC`, 2 sinA cos B + 2 sin B cos C + 2 sin cos A = sin A + sin B + sin C, then `Delta ABC`is

Options

A. isosceles
B. right angled
C. acute angled
D. none of these

Explanation

To solve the problem, we need to analyze the equation given in the triangle ABC:

\[ 2 \sin A \cos B + 2 \sin B \cos C + 2 \sin C \cos A = \sin A + \sin B + \sin C \]

### Step 1: Rewrite the left-hand side using the sine addition formula
We can use the sine addition formula, which states that \(2 \sin x \cos y = \sin(x+y) + \sin(x-y)\).

Applying this to each term:

1. \(2 \sin A \cos B = \sin(A + B) + \sin(A - B)\)
2. \(2 \sin B \cos C = \sin(B + C) + \sin(B - C)\)
3. \(2 \sin C \cos A = \sin(C + A) + \sin(C - A)\)

Thus, we can rewrite the left-hand side as:

\[ \sin(A + B) + \sin(A - B) + \sin(B + C) + \sin(B - C) + \sin(C + A) + \sin(C - A) \]

### Step 2: Substitute \(A + B + C = 180^\circ\)
Since in triangle ABC, \(A + B + C = 180^\circ\), we can express \(A + B\), \(B + C\), and \(C + A\) in terms of \(C\), \(A\), and \(B\) respectively:

- \(A + B = 180^\circ - C\) which gives \(\sin(A + B) = \sin C\)
- \(B + C = 180^\circ - A\) which gives \(\sin(B + C) = \sin A\)
- \(C + A = 180^\circ - B\) which gives \(\sin(C + A) = \sin B\)

Thus, we can simplify the left-hand side to:

\[ \sin C + \sin A + \sin B + \sin(A - B) + \sin(B - C) + \sin(C - A) \]

### Step 3: Set the equation
Now, we can set the equation:

\[ \sin C + \sin A + \sin B + \sin(A - B) + \sin(B - C) + \sin(C - A) = \sin A + \sin B + \sin C \]

### Step 4: Simplify the equation
Subtract \(\sin A + \sin B + \sin C\) from both sides:

\[ \sin(A - B) + \sin(B - C) + \sin(C - A) = 0 \]

### Step 5: Analyze the sine terms
The equation \(\sin(A - B) + \sin(B - C) + \sin(C - A) = 0\) implies that the angles \(A\), \(B\), and \(C\) are such that:

1. \(A - B = 0\) or \(B - C = 0\) or \(C - A = 0\)
2. This means \(A = B\), \(B = C\), or \(C = A\).

### Conclusion
Since we have established that two angles are equal, triangle ABC must be an **isosceles triangle**.