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प्रश्न
With the usual notations prove that `2{asin^2 "C"/(2) + "c"sin^2 "A"/(2)}` = a – b + c.
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उत्तर १
L.H.S. = `2{a sin^2 "C"/2 + "c"sin^2 "A"/2}`
= `2 {a ((1 – cos "C")/2) + c(1 – cos A/2)}`
= `2{(a - a cos "C" + c - c cos"A")}`
= `a + c - (c cos "A" + a cos "C")` `...[∵ "By Projection Rule b" = c cos"A" + a cos "C"]`
= a + c − b
= a - b + c
L. H. S. = R.H.S.
उत्तर २
L.H.S. = `2{a sin^2 "C"/2 + "c"sin^2 "A"/2}`
= `a(2sin^2 "C"/2)+c(2sin^2 "A"/2)`
= a (1 - cos C) + c (1 - cos A)
= `a[1 - (a^2+b^2-c^2)/(2ab)]+c[(b^2+c^2-a^2)/(2bc)]` ...[By cosine rule]
= `a[(2ab-a^2-b^2+c^2)/(2ab)]+c[(2bc-b^2-c^2+a^2)/(2bc)]`
= `(2ab-a^2-b^2+c^2)/(2b)+(2bc-b^2-c^2+a^2)/(2b)`
= `(2ab-a^2-b^2+c^2+2bc-b^2-c^2+a^2)/(2b)`
= `(2ab-2b^2+2bc)/(2b)`
= a - b + c
= RHS
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