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In a triangle ABC , with usual notations...

In a `triangle ABC` , with usual notations , prove that `(a-bcosC)/(b-acosC)=(cosB)/(cosA)`.

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Verified by Experts

Consider L.H.S `=(a-bcosC)/(b-cosC)`
But `a=c cosB+bcosC`
`b=c cosA+acosC`
`=(c cosB+bcosC-bcosC)/(c cosA+acosC-acosC)`
`=(c cosB)/(c cosA)`
`=(cosB)/(cosA)=R.H.S`.
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