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If in a triangle ABC, /C=60^@, then prov...

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

A

`(1)/(a+b+c)`

B

`(2)/(a+b+c)`

C

`(3)/(a+b+c)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
We have , `cosC=(a^(2)+b^(2)-c^(2))/(2ab)`
`rArrcos60^(@)=(a^(2)+b^(2)-c^(2))/(2ab)`
`rArra^(2)+b^(2)-c^(2)=ab`
`rArrb^(2)+bc+a^(2)+ac=ab+ac+bc+c^(2)`
`rArrb(b+c)+(a(a+c)=(a+c)(b+c)` On dividing by (a+c)(b+c) and then add 2 both sides we get
`1+(b)/(a+c)+1+(a)/(b+c)=3`
`rArr(1)/(a+c)+(1)/(b+c)=(3)/(a=b+c)`
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