Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

So by considering the given condition, we get

a², b², c² are in A.P

Then

b² - a² = c² - b² (this is the condition for A.P)

Substituting the values from equation (i), we get

⇒ (k sin B)² - (k sin A)² = (k sin C)² - (k sin B)²

⇒ k² (sin² B - sin² A) = k² (sin² C - sin² B)

⇒ sin (B + A) sin (B - A) = sin (C + B) sin (C - B)

(∵ sin²A - sin²B = sin (A + B) sin (A - B))

⇒ sin (π - C) sin (B - A) = sin (π - A) sin (C - B) (∵ π = A + B + C)

⇒ sin (C) sin (B - A) = sin (A) sin (C - B) (∵ sin (π - θ) = sin θ )

Shuffling this, we get

Canceling the like terms we get

But , so the above equation becomes,

⇒ cot A - cot B = cot B - cot C

Hence cot A, cot B, cot C are in AP

Hence proved