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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`
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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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