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In a ∆ABC, prove that (a2 – b2 + c2) tan B = (a2 + b2 – c2) tan C - Mathematics

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प्रश्न

In a ∆ABC, prove that (a2 – b2 + c2) tan B = (a2 + b2 – c2) tan C

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उत्तर

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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Properties of Triangle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.9 [पृष्ठ १४३]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.9 | Q 8 | पृष्ठ १४३

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