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Question
In , ΔABC with usual notations prove that
(a-b)2 cos2 `("C"/2) +("a"+"b")^2 "sin"^2("C"/2) = "c"^2`
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Solution
Taking LHS
= `("a"-"b")^2 "cos"^2 "C"/2 + ("a"+"b")^2 "sin"^2 "C"/2`
`=("a"^2+"b"^2 -2"ab") "cos"^2"C"/2 +("a"^2 +"b"^2 +2"ab")."sin""C"/2`
`= ("a"^2+"b"^2)"cos""C"/2 - 2 "ab" "cos"^2"C"/2 + ("a"^2+"b"^2) . "sin"^2"C" /2 + 2 "ab" "sin"^2 "C"/2`
`=("a"^2 + "b"^2)("cos"^2"C"/2 +"sin"^2"C"/2) - 2"ab"("cos"^2"C"/2 - "sin"^2"C"/2)`
=`("a"^2 +"b"^2) -2"ab" "cos""C"` {By cosine Rule}
= c2
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