The sides of a triangle are in AP. If the angles A and C are the greatest and smallest angle respectively, then `4 (1- cos A) (1-cos C)` is equal to

Options

A. a) `cos A - cos C `
B. b) `cos A cos C`
C. c) `cos A + cos C`
D. d) `cos C - cos A `

Answer

To solve the problem step by step, let's denote the sides of the triangle as \( a, b, c \) such that \( a < b < c \) (since angles A and C are the greatest and smallest angles respectively). Given that the sides are in arithmetic progression (AP), we can express them as:

1. Let the sides be \( a = x - d \), \( b = x \), and \( c = x + d \) for some \( x \) and \( d \).

### Step 1: Use the Law of Cosines
Using the Law of Cosines, we can express \( \cos A \) and \( \cos C \) in terms of the sides of the triangle.

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

### Step 2: Substitute the sides into the cosine formulas
Substituting \( a, b, c \) into the formulas:

\[ \cos A = \frac{x^2 + (x + d)^2 - (x - d)^2}{2x(x + d)} \]
\[ = \frac{x^2 + (x^2 + 2xd + d^2) - (x^2 - 2xd + d^2)}{2x(x + d)} \]
\[ = \frac{2x^2 + 2xd}{2x(x + d)} = \frac{x + d}{x + d} = 1 \]

\[ \cos C = \frac{(x - d)^2 + x^2 - (x + d)^2}{2(x - d)x} \]
\[ = \frac{(x^2 - 2xd + d^2) + x^2 - (x^2 + 2xd + d^2)}{2(x - d)x} \]
\[ = \frac{2x^2 - 4xd}{2(x - d)x} = \frac{x - 2d}{x - d} \]

### Step 3: Find \( 4(1 - \cos A)(1 - \cos C) \)
Now we need to compute \( 4(1 - \cos A)(1 - \cos C) \):

\[ 1 - \cos A = 1 - 1 = 0 \]
\[ 1 - \cos C = 1 - \frac{x - 2d}{x - d} = \frac{d}{x - d} \]

Thus,

\[ 4(1 - \cos A)(1 - \cos C) = 4(0)(\frac{d}{x - d}) = 0 \]

### Conclusion
The expression \( 4(1 - \cos A)(1 - \cos C) \) simplifies to 0.

### Final Answer
Thus, the value of \( 4(1 - \cos A)(1 - \cos C) \) is **0**.

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