If in a triangle `A B C ,/_C=60^0,` then prove that `1/(a+c)+1/(b+c)=3/(a+b+c)` .

Explanation

To prove that in triangle ABC, where angle C = 60°, the equation \( \frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c} \) holds, we will follow these steps:

### Step 1: Use the Cosine Rule
We start by applying the cosine rule to triangle ABC. The cosine rule states that:
\[ c^2 = a^2 + b^2 - 2ab \cos C \]
Given that \( C = 60^\circ \), we have \( \cos 60^\circ = \frac{1}{2} \). Thus, substituting this into the cosine rule gives:
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