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In any ΔABC, prove the following: cb cos Abc cos Acos Bcos Cc-b cos Ab-c cos A=cos Bcos C - Mathematics and Statistics

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Question

In any ΔABC, prove the following:

`("c" - "b cos A")/("b" - "c cos A") = ("cos B")/("cos C")`

Sum
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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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Chapter 3: Trigonometric Functions - Miscellaneous exercise 3 [Page 109]

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Balbharati Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 11.2 | Page 109

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