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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

In any Δ ABC, prove the following: cos 2Aa2-cos 2Bb2=1a2-1b2

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प्रश्न

In any Δ ABC, prove the following:

`"cos 2A"/"a"^2 - "cos 2B"/"b"^2 = 1/"a"^2 - 1/"b"^2`

बेरीज
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उत्तर

By sine rule,

`"sin A"/"a" = "sin B"/"b"`

∴ `("sin"^2"A")/"a"^2 = ("sin"^2"B")/"b"^2`   ....(1)

LHS = `"cos 2A"/"a"^2 - "cos 2B"/"b"^2`

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

`= 1/"a"^2 - (2 "sin"^2 "A")/"a"^2 - 1/"b"^2 + (2 "sin"^2 "B")/"b"^2`

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

`= 1/"a"^2 - 1/"b"^2 - 2(("sin"^2"B")/"b"^2 - ("sin"^2"B")/"b"^2)`    .....[By (1)]

`= 1/"a"^2 - 1/"b"^2 - 2 xx 0`

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

= RHS

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पाठ 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
पाठ 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 11.6 | पृष्ठ १०९

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