Q. 27

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

Answer :

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


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