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प्रश्न
In ΔABC, prove that `("a - b")^2 cos^2 "C"/2 + ("a + b")^2 sin^2 "C"/2 = "c"^2`
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उत्तर
LHS = `("a - b")^2 cos^2 "C"/2 + ("a + b")^2 sin^2 "C"/2 = "c"^2`
`= ("a"^2 + "b"^2 - 2"ab")cos^2 "C"/2 + ("a"^2 + "b"^2 + 2"ab")sin^2 "C"/2`
`= ("a"^2 + "b"^2)cos^2 "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)`
= a2 + b2 - 2ab cos C
= c2 = RHS
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