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Question
In ∆ABC, if ∠A = `pi/2`, then prove that sin(B − C) = `("b"^2 - "c"^2)/("b"^2 + "c"^2)`
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Solution
In ∆ABC, ∠A = `pi/2` .......[Given]
∴ sin A = `sin pi/2` = 1
and A + B + C = π
∴ B + C = `pi/2`
∴ B = `pi/2 - "C"` and C = `pi/2 - "B"`
∴ sin B = `sin (pi/2 - "C")`
= cos C and sin C
= `sin (pi/2 - "B")`
= cos B .......(i)
In ∆ABC by sine rule, we have
`"a"/(sin "A") = "b"/(sin "B") = "c"/(sin "C")`
∴ `"a"/1 = "b"/(sin "B") = "c"/(sin "C")`
∴ b = a sin B, c = a sin C .......(ii)
∴ R.H.S. = `("b"^2 - "c"^2)/("b"^2 + "c"^2)`
= `("a"^2sin^2"B" - "a"^2sin^2"C")/("a"^2sin^2"B" + "a"^2sin^2"C")` .......[From (ii)]
= `(sin^2"B" - sin^2"C")/(sin^2"B" + sin^2"C")`
= `(sin"B"*(sin"B") - sin"C"*(sin"C"))/(sin^2"B" + sin^2"C")`
= `(sin"B" cos"C" - sin"C" cos"B")/(sin^2"B" + cos^2"B")` .......[From (i)]
= `(sin("B" - "C"))/1`
= sin (B − C)
= L.H.S.
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