Given: In triangle ABC,
sinA/ sinC = sin(A-B ) / sin (B-C)

To prove: a^2, b^2, c^2 are in AP (Arithmetic Progression)

Proof:

1. Using the sine rule:
In triangle ABC, the sine rule states:
a/sinA = b/sinB = c/sinC

2. Rearranging the given relation:
sinA / sinC = sin(A-B) / sin(B-C)

Using the sine rule, we can rewrite the given relation as:
(a/sinA) / (c/sinC) = (sin(A-B) / sinB) / (sin(B-C) / sinC)

Simplifying further, we get:
(a/c) = (sin(A-B) / sinB) / (sin(B-C) / sinC)

3. Applying cross multiplication:
(a/c) = [(sinAcosB - cosAsinB) / sinB] / [(sinBcosC - cosBsinC) / sinC]

Simplifying further, we get:
(a/c) = [(sinAcosB - cosAsinB) / sinB] * [sinC / (sinBcosC - cosBsinC)]

4. Simplifying the numerator:
Using the trigonometric identity sin(A-B) = sinAcosB - cosAsinB, we can rewrite the numerator as:
sin(A-B)

5. Simplifying the denominator:
Using the trigonometric identity sin(B-C) = sinBcosC - cosBsinC, we can rewrite the denominator as:
sin(B-C)

6. Final relation:
(a/c) = sin(A-B) / sin(B-C)

7. Applying the sine rule:
From the sine rule, we know that a/sinA = b/sinB = c/sinC

Therefore, we can rewrite the relation (a/c) = sin(A-B) / sin(B-C) as:
(b/sinB) / (c/sinC) = sin(A-B) / sin(B-C)

Simplifying further, we get:
(b/c) = sin(A-B) / sin(B-C)

8. Simplifying the numerator:
Using the trigonometric identity sin(A-B) = sinAcosB - cosAsinB, we can rewrite the numerator as:
sinAcosB - cosAsinB

9. Simplifying the denominator:
Using the trigonometric identity sin(B-C) = sinBcosC - cosBsinC, we can rewrite the denominator as:
sinBcosC - cosBsinC

10. Final relation:
(b/c) = (sinAcosB - cosAsinB) / (sinBcosC - cosBsinC)

11. Applying the sine rule:
From the sine rule, we know that a/sinA = b/sinB = c/sinC

Therefore, we can rewrite the relation (b/c) = (sinAcosB - cosAsinB) / (sinBcosC - cosBsinC) as:
(a