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

Explanation

By the cosine formula, we have
`c^(2) = a^(2) + b^(2) - 2ab cos C`
or `c^(2) = a^(2) + b^(2) - 2ab cos 60^(@) = a^(2) + b^(2) - ab` (i)
Now `(1)/(a + c) + (1)/(b + c) - (3)/(a + b + c)`
`=[((b + c) (a + b +c) + (a + c) (a + b+ c) -3 (a + c) (b + c))/((a + b) (b + c) (a + b + c))]`
`= ((a^(2) + b^(2) - ab) - c^(2))/((a + b) (b + c) (a + b+ c)) = 0` [from Eq. (i)]
or `(1)/(a + c) + (1)/(b + c) = (3)/(a + b + c)`