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In a triangle A B C ,/C=60^@, then prove...

In a triangle `A B C ,/_C=60^@,` then prove that: `1/(a+c)+1/(b+c)=3/(a+b+c)dot`

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Given`∠C=60^o`
We know that
`c^2=a^2+b^2−2abcosC`
`c^2=a^2+b^2−2abcos60^@`
or `c^2=a^2+b^2−ab ...........(1)`
Now given that
`1/(a+c)​+1/(b+c)​=3/(a+b+c)​`
solving both side equation we get
...
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