In any ∆ABC, if a2, b2, c2 are in A.P., prove that cot A, cot B, and cot C are also in A.P
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