In a triangle ABC the expression `acosBcosC+bcosC cosA+c cosA cosB` equals to :

Options

A. `(rs)/(R )`
B. `(r )/(sR)`
C. `(R )/(rs)`
D. `(Rs)/(r )`

Answer

To solve the expression \( A \cos B \cos C + B \cos C \cos A + C \cos A \cos B \) in terms of the inradius \( r \), semi-perimeter \( s \), and circumradius \( R \) of triangle \( ABC \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ A \cos B \cos C + B \cos C \cos A + C \cos A \cos B \]

### Step 2: Use Trigonometric Relationships
From the triangle, we can express \( B \) in terms of \( A \) and \( C \) using the cosine rule. We know that:
\[ B = C \cos A + A \cos C \]

### Step 3: Factor Out Common Terms
We can factor out \( \cos B \) from the first and third terms:
\[ = \cos B (A \cos C + C \cos A) + B \cos C \cos A \]

### Step 4: Substitute for \( A \cos C + C \cos A \)
Substituting \( A \cos C + C \cos A \) with \( b \):
\[ = B \cos B + b \cos C \cos A \]

### Step 5: Use the Sine Rule
Using the sine rule, we know:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
From this, we can express \( B \) in terms of \( R \):
\[ B = 2R \sin B \]

### Step 6: Substitute Back into the Expression
Substituting \( B \) into our expression gives:
\[ = 2R \sin B \cos B + b \cos C \cos A \]

### Step 7: Use Area of Triangle
Recall that the area \( \Delta \) of triangle \( ABC \) can be expressed as:
\[ \Delta = R \cdot \frac{a \cdot b \cdot c}{4R} \]
Thus, we can relate the area to \( R \) and the sines:
\[ \Delta = R^2 \sin A \sin B \sin C \]

### Step 8: Relate Area to Inradius and Semi-perimeter
We also know that:
\[ \Delta = r \cdot s \]
where \( s \) is the semi-perimeter of the triangle.

### Step 9: Final Expression
Combining the results, we find:
\[ A \cos B \cos C + B \cos C \cos A + C \cos A \cos B = \frac{rs}{R} \]

### Conclusion
The final result of the expression \( A \cos B \cos C + B \cos C \cos A + C \cos A \cos B \) is:
\[ \frac{rs}{R} \]