If `A+B+C=180^(@), " then " (cosAcosC+cos(A+B)cos(B+C))/(cosAsinC-sin(A+B)cos(B+C))` simplifies to :

Options

A. `-cot C`
B. 0
C. `tanC`
D. `cot C`

Answer

To simplify the expression \((\cos A \cos C + \cos(A+B) \cos(B+C)) / (\cos A \sin C - \sin(A+B) \cos(B+C))\) given that \(A + B + C = 180^\circ\), we can follow these steps:

### Step 1: Use the identity for angles
Since \(A + B + C = 180^\circ\), we can express \(C\) as \(C = 180^\circ - (A + B)\).

### Step 2: Substitute \(C\) in the expression
Now, substituting \(C\) in the expression:
- \(\cos C = \cos(180^\circ - (A + B)) = -\cos(A + B)\)
- \(\sin C = \sin(180^\circ - (A + B)) = \sin(A + B)\)

### Step 3: Rewrite the expression
Now, we can rewrite the expression:
\[ \frac{\cos A (-\cos(A + B)) + \cos(A + B) \cos(B + (180^\circ - (A + B)))}{\cos A \sin(A + B) - \sin(A + B) \cos(B + (180^\circ - (A + B)))} \]
This simplifies to:
\[ \frac{-\cos A \cos(A + B) + \cos(A + B)(-\cos A)}{\cos A \sin(A + B) - \sin(A + B)(-\cos A)} \]

### Step 4: Simplify the numerator
The numerator simplifies to:
\[ -\cos A \cos(A + B) - \cos A \cos(A + B) = -2\cos A \cos(A + B) \]

### Step 5: Simplify the denominator
The denominator simplifies to:
\[ \cos A \sin(A + B) + \sin(A + B) \cos A = 2\cos A \sin(A + B) \]

### Step 6: Combine the results
Now, we have:
\[ \frac{-2 \cos A \cos(A + B)}{2 \cos A \sin(A + B)} = -\frac{\cos(A + B)}{\sin(A + B)} \]

### Step 7: Final simplification
This simplifies to:
\[ -\cot(A + B) \]

### Step 8: Use the identity for \(A + B\)
Since \(A + B = 180^\circ - C\), we can express:
\[ -\cot(180^\circ - C) = \cot C \]

### Final Answer
Thus, the expression simplifies to \(\cot C\).

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