Q. 27

# In any ∆ABC, if a^{2}, b^{2}, c^{2} 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

a^{2}, b^{2}, c^{2} are in A.P

Then

b^{2} - a^{2} = c^{2} - b^{2} (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}

⇒ k^{2} (sin^{2} B - sin^{2} A) = k^{2} (sin^{2} C - sin^{2} B)

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

(∵ sin^{2}A - sin^{2}B = 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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