In any ΔABC, prove that a2 sin ( B – C ) = (b2 – c2) sin A

1.	In any ΔABC, prove that a2 sin ( B – C ) = (b2 – c2) sin A
Answer» To prove: a² sin ( B – C ) = (b² – c²) sin A We know that, a/sinA = b/sinB = c/sinC = 2R where R is the circumradius. Therefore, a = 2R sinA ---- (a) Similarly, b = 2R sinB and c = 2R sinC From Right hand side, = (b² – c²) sin A = {(2R sinB)² – (2R sinC)² } sinA = 4R2 ( sin²B – sin²C )sinA We know, sin²B – sin²C = sin(B + C)sin(B – C) So, = 4R² (sin(B + C)sin(B – C))sinA = 4R² (sin(π – A)sin(B – C))sinA [ As, A + B + C = π] = 4R² (sinAsin(B – C))sinA [ As, sin(π − θ) = sin θ ] = 4R²sin²A sin(B – C) = a 2sin(B – C) [From (a)] = Left hand side. [Proved]

Answer»

To prove: a² sin ( B – C ) = (b² – c²) sin A

We know that, a/sinA = b/sinB = c/sinC = 2R where R is the circumradius.

Therefore,

a = 2R sinA ---- (a)

Similarly, b = 2R sinB and c = 2R sinC

From Right hand side,

= (b² – c²) sin A

= {(2R sinB)² – (2R sinC)² } sinA

= 4R2 ( sin²B – sin²C )sinA

We know, sin²B – sin²C = sin(B + C)sin(B – C)

So,

= 4R² (sin(B + C)sin(B – C))sinA

= 4R² (sin(π – A)sin(B – C))sinA [ As, A + B + C = π]

= 4R² (sinAsin(B – C))sinA [ As, sin(π − θ) = sin θ ]

= 4R²sin²A sin(B – C)

= a 2sin(B – C) [From (a)]

= Left hand side. [Proved]

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