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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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Question

With the usual notations, show that
(c2 − a2 + b2) tan A = (a2 − b2 + c2) tan B = (b2 − c2 + a2) tan C

Sum
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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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Chapter 3: Trigonometric Functions - Miscellaneous exercise 3 [Page 110]

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Balbharati Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 17 | Page 110

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