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In any Δ ABC, prove the following: ac cos B - bc cos A = a2 - b2

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प्रश्न

In any Δ ABC, prove the following:

ac cos B - bc cos A = a2 - b2

योग
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उत्तर

LHS = ac cos B - bc cos A = a2 - b

`= "ac"(("c"^2 + "a"^2 - "b"^2)/"2ca") - "bc"(("b"^2 + "c"^2 - "a"^2)/"2bc")`

`= 1/2 ("c"^2 + "a"^2 - "b"^2) - 1/2 ("b"^2 + "c"^2 - "a"^2)`

`= 1/2 ("c"^2 + "a"^2 - "b"^2 - "b"^2 - "c"^2 + "a"^2)`

`= 1/2 (2"a"^2 - 2"b"^2)`

`= "a"^2 - "b"^2`

= RHS

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometric Functions - Miscellaneous exercise 3 [पृष्ठ १०९]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 11.4 | पृष्ठ १०९

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