# In ∆ABC, if ∠A = π2, then prove that sin(B − C) = b2-c2b2+c2 - Mathematics and Statistics

Sum

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.

Concept: Solutions of Triangle
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