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प्रश्न
In ∆ABC, prove that `(cos 2"A")/"a"^2 - (cos 2"c")/"c"^2 = 1/"a"^2 - 1/"c"^2`
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उत्तर
Consider L.H.S. = `(cos 2"A")/"a"^2 - (cos 2"c")/"c"^2`
= `(1 - 2sin^2 "A")/"a"^2 - (1 - 2sin^2"C")/"c"^2`
= `1/"a"^2 - 2 (sin^2"A")/"a"^2 - 1/"c"^2 + 2 (sin^2"C")/"c"^2`
= `1/"a"^2 - 2"k"^2 - 1/"c"^2 + 2"k"^2` .......[By since rule]
= `1/"a"^2 - 1/"c"^2`
= R.H.S.
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