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प्रश्न
In ΔABC, prove that `("b"^2 - "c"^2)/"a" cos"A" + ("c"^2 - "a"^2)/"b" cos"B" + ("a"^2 - "b"^2)/"c" cos "C"` = 0
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उत्तर
In ΔABC by cosine rule, we have
cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`, cosB = `("a"^2 + "c"^2 - "b"^2)/(2"ac")`, cos C = `("a"^2 + "b"^2 - "c"^2)/(2"ab")`
∴ L.H.S. = `("b"^2 - "c"^2)/"a" cos"A" + ("c"^2 - "a"^2)/"b" cos"B" + ("a"^2 - "b"^2)/"c" cos "C"`
= `("b"^2 - "c"^2)/"a" xx ("b"^2 + "c"^2 - "a"^2)/(2"bc") + ("c"^2 - "a"^2)/"b" xx ("a"^2 + "c"^2 - "b"^2)/(2"ac") + ("a"^2 - "b"^2)/"c" xx ("a"^2 + "b"^2 - "c"^2)/(2"ab")`
= `1/(2"abc") ("b"^4 + "b"^2"c"^2 - "a"^2"b"^2 - "b"^2"c"^2 - "c"^4 + "a"^2"c"^2 + "a"^2"c"^2 + "c"^4 - "b"^2"c"^2 - "a"^4 - "a"^2"c"^2 + "a"^2"b"^2 + "a"^4 + "a"^2"b"^2 - "b"^4 + "b"^2"c"^2)`
= 0
= R.H.S
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