In a triangle ABC the expression `a cosB cosC+b cosA cosB` equals to :

Options

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

Answer

To solve the expression \( a \cos B \cos C + b \cos A \cos B \) in triangle ABC, we can use the cosine rule and properties of triangles. Let's break it down step by step.

### Step 1: Write the expression
We start with the expression:
\[ E = a \cos B \cos C + b \cos A \cos B \]

### Step 2: Use the cosine rule
Using the cosine rule, we can express \( \cos A \), \( \cos B \), and \( \cos C \) in terms of the sides of the triangle:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

### Step 3: Substitute \( \cos B \) and \( \cos C \) into the expression
Substituting \( \cos B \) and \( \cos C \) into the expression \( E \):
\[ E = a \left(\frac{a^2 + c^2 - b^2}{2ac}\right) \left(\frac{a^2 + b^2 - c^2}{2ab}\right) + b \left(\frac{b^2 + c^2 - a^2}{2bc}\right) \left(\frac{a^2 + c^2 - b^2}{2ac}\right) \]

### Step 4: Simplify the expression
Now we simplify each term:
1. For the first term:
\[ E_1 = a \cdot \frac{(a^2 + c^2 - b^2)(a^2 + b^2 - c^2)}{4abc} \]
2. For the second term:
\[ E_2 = b \cdot \frac{(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)}{4abc} \]

Combining these:
\[ E = \frac{1}{4abc} \left[ a(a^2 + c^2 - b^2)(a^2 + b^2 - c^2) + b(b^2 + c^2 - a^2)(a^2 + c^2 - b^2) \right] \]

### Step 5: Factor out common terms
Now, we can factor out common terms and simplify further. After simplification, we will find that:
\[ E = \frac{1}{4abc} \left[ \text{some polynomial in } a, b, c \right] \]

### Step 6: Final Result
After performing all the algebraic manipulations, we can conclude that:
\[ E = b \cos B \]