In any triangle ABC, prove the: b sin B – c sin C = a sin (B – C)

1.	In any triangle ABC, prove the: b sin B – c sin C = a sin (B – C)
Answer» On using the sine rule we know, a/sin A = b/sin B = c/sin C = k c/sin C = k Therefore, c = k sin C Now similarly, a = k sin A And b = k sin B As we know, Now let us consider the LHS b sin B – c sin C By substituting the values of b and c in the above equation, we get k sin B sin B – k sin C sin C = k (sin² B – sin² C) ……….(i) As we know, Sin² B – sin² C = sin (B + C) sin (B – C), By substituting the above values in equation (i), we get k (sin² B – sin² C) = k (sin (B + C) sin (B – C)) [since, π = A + B + C ⇒ B + C = π –A] Now the above equation becomes, = k (sin (π –A) sin (B – C)) [since, sin (π – θ) = sin θ] = k (sin (A) sin (B – C)) From sine rule, a = k sin A, therefore the above equation becomes, = a sin (B – C) = RHS Thus proved.

Answer»

On using the sine rule we know,

a/sin A = b/sin B = c/sin C = k

c/sin C = k

Therefore, c = k sin C

Now similarly, a = k sin A

And b = k sin B

As we know,

Now let us consider the LHS

b sin B – c sin C

By substituting the values of b and c in the above equation, we get

k sin B sin B – k sin C sin C = k (sin² B – sin² C) ……….(i)

As we know,

Sin² B – sin² C = sin (B + C) sin (B – C),

By substituting the above values in equation (i), we get

k (sin² B – sin² C) = k (sin (B + C) sin (B – C)) [since, π = A + B + C ⇒ B + C = π –A]

Now the above equation becomes,

= k (sin (π –A) sin (B – C)) [since, sin (π – θ) = sin θ]

= k (sin (A) sin (B – C))

From sine rule, a = k sin A, therefore the above equation becomes,

= a sin (B – C)

= RHS

Thus proved.

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