Question -
Answer -
By using the sine rule we know,
So, c = k sin C
Similarly, a = k sin A
And b = k sin B
We know,
Now let us consider RHS:
(b2 – c2) sin A …
Substituting the values of b and c in the aboveequation, we get
(b2 – c2) sin A = [(k sinB)2 – ( k sin C)2] sin A
= k2 (sin2 B – sin2 C)sin A………. (i)
We know,
Sin2 B – sin2 C = sin(B + C) sin (B – C),
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]
Rearranging the above equation we get
= (k sin (A))( sin (B – C)) (k sin A)
From sine rule, a = k sin A, so the above equationbecomes,
= a2 sin (B – C)
= RHS
Hence proved.