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In a Δ ABC, with usual notations prove that:` (a -bcos C) /(b -a cos C )= cos B/ cos A`

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#### Solution

Consider L.H.S. `=(a-bcosC)/(b-acosC)`

`=(b cosC+c cosB-b cosC)/(a cosC+c cosA-a cosC)`

`=(c cosB)/(c cosA) `

`=cosB/cosA=R.H.S`

`therefore (a-bcosC)/(b-acosC)=(cosB)/(cosA)`

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