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Question
With the usual notations, show that
(c2 − a2 + b2) tan A = (a2 − b2 + c2) tan B = (b2 − c2 + a2) tan C
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Solution
By sine rule,
`"a"/("sin A") = "b"/("sin B") = "c"/("sin C")` = k
`"a"/("sin A") = "k", "b"/("sin B") = "k", "c"/("sin C")` = k
∴ sin A = `"a"/"k"`, sin B = `"b"/"k"`, sin C = `"c"/"k"`
Now,
(c2 − a2 + b2) tan A = (c2 − a2 + b2). `"sin A"/"cos A"`
`= ("c"^2 + "b"^2 - "a"^2) xx ("k"/"a")/((("c"^2 + "b"^2 - "a"^2)/"2bc"))`
`= ("c"^2 + "b"^2 - "a"^2)/"k" xx "2abc"/("c"^2 + "b"^2 - "a"^2)`
= `("2abc")/"k"` .....(1)
(a2 - b2 + c2) tan B = (a2 - b2 + c2) .`"sin B"/"cos B"`
`= ("a"^2 + "c"^2 - "b"^2) xx ("k"/"b")/((("a"^2 + "c"^2 - "b"^2)/"2ac"))`
`= ("a"^2 + "c"^2 - "b"^2)/"k" xx "2abc"/("a"^2 + "c"^2 - "b"^2)`
= `("2abc")/"k"` ....(2)
(b2 − c2 + a2) tan C = (b2 − c2 + a2). `"sin C"/"cos C"`
`= ("a"^2 + "b"^2 - "c"^2) xx ("k"/"c")/((("a"^2 + "b"^2 - "c"^2)/"2ab"))`
`= ("a"^2 + "b"^2 - "c"^2)/"k" xx "2abc"/("a"^2 + "b"^2 - "c"^2)`
= `("2abc")/"k"` .....(3)
From (1), (2) and (3), we get
(c2 − a2 + b2) tan A = (a2 − b2 + c2) tan B = (b2 − c2 + a2) tan C
`("2abc")/"k" = ("2abc")/"k" = ("2abc")/"k"`
1 = 1= 1
All are equals.
L. H. S. = R. H. S.
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