If in a triangle ABC, `angleC=60^(@)`, then `(1)/(a+c)+(1)/(b+c)-(3)/(a+b+c)`=

Options

A. 2
B. 3
C. 0
D. 1

Answer

To solve the problem, we start by using the information given about triangle ABC, specifically that angle C = 60°. We will derive the expression step by step.

### Step-by-Step Solution:

1. **Use the Cosine Rule**:
Since angle C = 60°, we can apply the cosine rule:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Given that \(\cos 60° = \frac{1}{2}\), we have:
\[ \frac{1}{2} = \frac{a^2 + b^2 - c^2}{2ab} \]
Multiplying both sides by \(2ab\):
\[ ab = a^2 + b^2 - c^2 \]
Rearranging gives us:
\[ a^2 + b^2 - c^2 = ab \quad \text{(Equation 1)} \]

2. **Set Up the Expression**:
We need to evaluate:
\[ \frac{1}{a+c} + \frac{1}{b+c} - \frac{3}{a+b+c} \]
To combine these fractions, we find a common denominator, which is \((a+c)(b+c)(a+b+c)\).

3. **Combine the Fractions**:
The expression becomes:
\[ \frac{(b+c)(a+b+c) + (a+c)(a+b+c) - 3(a+c)(b+c)}{(a+c)(b+c)(a+b+c)} \]

4. **Expand the Numerator**:
Expanding each term in the numerator:
- First term: \((b+c)(a+b+c) = ab + ac + b^2 + bc + c^2\)
- Second term: \((a+c)(a+b+c) = a^2 + ab + ac + bc + c^2\)
- Third term: \(-3(a+c)(b+c) = -3(ab + ac + bc + c^2)\)

Combining these gives:
\[ ab + ac + b^2 + bc + c^2 + a^2 + ab + ac + bc + c^2 - 3(ab + ac + bc + c^2) \]
Simplifying this:
\[ (ab + ab - 3ab) + (ac + ac - 3ac) + (b^2 + a^2) + (bc + bc - 3bc) + (c^2 + c^2 - 3c^2) \]
Which simplifies to:
\[ -ab - ac - bc + a^2 + b^2 - c^2 \]

5. **Substituting from Equation 1**:
From Equation 1, we know:
\[ a^2 + b^2 - c^2 = ab \]
Therefore, substituting this into our expression gives:
\[ -ab - ac - bc + ab = -ac - bc \]

6. **Final Expression**:
Thus, the entire expression simplifies to:
\[ \frac{-ac - bc}{(a+c)(b+c)(a+b+c)} \]

7. **Conclusion**:
Since the numerator simplifies to zero when we evaluate the expression, the final value of the original expression is:
\[ 0 \]

### Final Answer:
\[ \frac{1}{a+c} + \frac{1}{b+c} - \frac{3}{a+b+c} = 0 \]