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In a triangle ABC, if sinAsin(B-C)=sinCs...

In a triangle ABC, if `sinAsin(B-C)=sinCsin(A-B),` then prove that `cotA ,cotB ,cotC` are in `AdotPdot`

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`sin(B-C)/sinC = sin(A-B)/sinA`
`=>(sinBcosC - sinCcosB)/sinC = (sinAcosB - sinBcosA)/sinA`
`=>sinBcotC - cosB = cosB - sinBcotA`
Diving both sides by `sinB`,
`=>cotC -cotB = cotB - cotA`
`=>2cotB = cotA+cotC`
Therefore, `cotA,cotB and cotC` are in A.P.
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