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Question
In any ΔABC, prove the following:
`("c" - "b cos A")/("b" - "c cos A") = ("cos B")/("cos C")`
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Solution
L.H.S. = `("c" - "b cos A")/("b" - "c cos A")`
`= ("c" - "b" (("b"^2 + "c"^2 - "a"^2)/(2"bc")))/("b" - "c" (("b"^2 + "c"^2 - "a"^2)/(2"bc")))`
`= ("c" - (("b"^2 + "c"^2 - "a"^2)/"2c"))/("b" - (("b"^2 + "c"^2 - "a"^2)/"2b"))`
`= ((2"c"^2 - "b"^2 - "c"^2 + "a"^2)/"2c")/((2"b"^2 - "b"^2 - "c"^2 + "a"^2)/"2b")`
= `((("c"^2 + "a"^2 - "b"^2)/"2c"))/((("a"^2 + "b"^2 - "c"^2)/"2b")) `
`= ((("c"^2 + "a"^2 - "b"^2)/"2ca"))/((("a"^2 + "b"^2 - "c"^2)/"2ab")`
= `"cos B"/"cos C"`
= R.H.S.
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