In any triangle ABC, prove the: a2 sin (B – C) = (b2 – c2) sin

1.	In any triangle ABC, prove the: a2 sin (B – C) = (b2 – c2) sin A
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 Similarly, a = k sin A And b = k sin B As we know, Now let us consider the RHS (b² – c²) sin A By substituting the values of b and c in the above equation, we get (b² – c²) sin A = [(k sin B)² – ( k sin C)²] sin A = k²(sin² B – sin² C) sin A………. (i) As we know, Sin² B – sin² C = sin (B + C) sin (B – C), By substituting the above values in equation (i), we get = k²(sin (B + C) sin (B – C)) sin A [since, π = A + B + C ⇒ B + C = π –A] = k²(sin (π –A) sin (B – C)) sin A = k²(sin (A) sin (B – C)) sin A [since, sin (π – θ) = sin θ] Then, rearranging the above equation we get = (k sin (A))( sin (B – C)) (k sin A) From sine rule, a = k sin A, therefore the above equation becomes, = a² sin (B – C) = RHS Thus proved.

In any triangle ABC, prove the: a2 sin (B – C) = (b2 – c2) sin A

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

Similarly, a = k sin A

And b = k sin B

As we know,

Now let us consider the RHS

(b² – c²) sin A

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

(b² – c²) sin A = [(k sin B)² – ( k sin C)²] sin A

= k²(sin² B – sin² C) sin A………. (i)

As we know,

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

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

= k²(sin (B + C) sin (B – C)) sin A [since, π = A + B + C ⇒ B + C = π –A]

= k²(sin (π –A) sin (B – C)) sin A

= k²(sin (A) sin (B – C)) sin A [since, sin (π – θ) = sin θ]

Then, rearranging the above equation we get

= (k sin (A))( sin (B – C)) (k sin A)

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

= a² sin (B – C)

= RHS

Thus proved.