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In , Abc Prove that

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Question

 In , ΔABC prove that 

`"sin"(("B" - "C")/2) = (("b" - "c")/"a") "cos"("A"/2)`                               

Sum
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Solution

RHS = `(("b"-"c")/"a") "cos""A"/2`         ...(by sine rule)

=`(("k" "sin" "B" - "k" "sin" "C")/("k" "sin""A")) . "cos""A"/2`

= `["k"["sin""B" - "sin""C"]]/("k""sin""A") ."cos""A"/2`

=`[[2"cos" ("B"+"C")/2 . "sin" ("B" -"C")/2]]/(2 "sin" "A"/2 "cos""A"/2) . "cos""A"/2` 

= `(2"cos" (pi/2-"A"/2) "sin"
(("B"-"C")/2))/(2 "sin" "A"/2)`

=` "sin"(("B"-"C")/2) = "LHS"`

Hence, the required result is proved.                           

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2018-2019 (March) Set 1

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