In ∆ABC, if ∠A = `pi/2`, then prove that sin(B − C) = `("b"^2 - "c"^2)/("b"^2 + "c"^2)`

#### 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.