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Question
In a ∆ABC, prove that (a2 – b2 + c2) tan B = (a2 + b2 – c2) tan C
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Solution
We know `"a"/sin"A" = "b"/sin "B" = "c"/sin "C"` = 2R
`"a"/sin"A"` = 2R ⇒ a = 2R sin A
`"b"/sin"B"` = 2R ⇒ b = 2R sin B
`"c"/sin"C"` = 2R ⇒ c = 2R sin C
Also we know cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`
cos B = `("c"^2 + "a"^2 - "b"^2)/(2"ca")`
cos C = `("a"^2 + "b"^2 - "c"^2)/(2"ab")`
`("a"^2 + "b"^2 - "c"^2)/("a"^2 - "b"^2 + "c"^2) = ("a"^2 + "b"^2 - "c"^2)/("c"^2 + "a"^2 + "b"^2)`
= `(("a"^2 + "b"^2 - "c"^2)/(2"abc"))/(("c"^2 + "a"^2 - "b"^2)/(2"abc"))`
= `(1/"c" xx ("a"^2 + "b"^2 - "c"^2)/(2"ab"))/(1/"b" xx ("c"^2 + "a"^2 - "b"^2)/(2"ca"))`
= `(1/"c" xx cos "C")/(1/"b" xx cos "B")`
= `("b" cos "C")/("c" cos "B")`
= `(2"R" sin"B" cos"C")/(2"R" sin"C" cos"B")`
`("a"^2 + "b"^2 - "c"^2)/("a"^2 - "b"^2 + "c"^2) = sin"B"/sin"C" * cos"C"/sin"C"`
`("a"^2 + "b"^2 - "c"^2)/("a"^2 - "b"^2 + "c"^2)` = tan B . cot C
`("a"^2 + "B"^2 - "c"^2)/cot"C"` = (a2 – b2 + c2) tan B
(a2 + b2 + c2) tan C = (a2 – b2 + c2) tan B
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