In $ riangle ABC$, if $\cot A$, $\cot B$, $\cot C$ are in A.P. then show that $a^2$, $b^2$, $c^2$ are also in A.P.

Question

In $	riangle ABC$, if $\cot A$, $\cot B$, $\cot C$ are in A.P. then show that $a^2$, $b^2$, $c^2$ are also in A.P.

Accepted Answer

$\cot A=\dfrac{\cos A}{\sin A}=\dfrac{(b^2+c^2-a^2)/2bc}{a/2R}=\dfrac{R(b^2+c^2-a^2)}{abc}$ , and similarly $\cot B=\dfrac{R(c^2+a^2-b^2)}{abc}$ , $\cot C=\dfrac{R(a^2+b^2-c^2)}{abc}$ . The A.P. condition $2\cot B=\cot A+\cot C$ becomes $2R(c^2+a^2-b^2)=R(b^2+c^2-a^2)+R(a^2+b^2-c^2)$ . The right side simplifies to $2Rb^2\cdot$ ... precisely: R.H.S. $=R[(b^2+c^2-a^2)+(a^2+b^2-c^2)]=R\cdot2b^2$ . So $2(c^2+a^2-b^2)=2b^2\implies c …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL2 marks
$\triangle ABC$
›Reveal solutionSolution

Substitute $\cos C$ and $\cos B$ from the cosine rule and simplify.

By the cosine rule:

$\cos C=\frac{a^2+b^2-c^2}{2ab}, \qquad \cos B=\frac{a^2+c^2-b^2}{2ac}$

LHS $=a(b\cos C-c\cos B) = ab\cos C - ac\cos B$

$ab\cos C = ab\cdot\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2-c^2}{2}$

$ac\cos B = ac\cdot\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+c^2-b^2}{2}$

…

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CBSE 2023Set ANNUAL2 marksMCQ
$\triangle ABC$
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{2}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

Compare with the cosine rule $b^2=a^2+c^2-2ac\cos B$ .

By the cosine rule, $b^2=a^2+c^2-2ac\cos B \Rightarrow a^2+c^2-b^2=2ac\cos B$ .

…

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CBSE 2022Set ANNUAL2 marksMCQ
$\triangle ABC$
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{2}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

Compare with the Cosine Rule formula for $\cos B$ .

By the Cosine Rule: $\cos B = \dfrac{a^2+c^2-b^2}{2ac}$

Given $c^2+a^2-b^2=ac$ : …

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CBSE 2017Set ANNUAL2 marks
$\triangle ABC$
›Reveal solutionSolution

Substitute $\cos B$ and $\cos C$ from the cosine rule and simplify.

By the cosine rule:

$\cos C = \dfrac{a^2+b^2-c^2}{2ab}, \qquad \cos B = \dfrac{a^2+c^2-b^2}{2ac}$

$a(b\cos C - c\cos B) = a\left[b\cdot\dfrac{a^2+b^2-c^2}{2ab} - c\cdot\dfrac{a^2+c^2-b^2}{2ac}\right]$

$= a\left[\dfrac{a^2+b^2-c^2}{2a} - \dfrac{a^2+c^2-b^2}{2a}\right] = \dfrac{1}{2}\Big[(a^2+b^2-c^2)-(a^2+c^2-b^2)\Big]$

…

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Q.In $\triangle ABC$ , if $\cot A$ , $\cot B$ , $\cot C$ are in A.P. then show that $a^2$ , $b^2$ , $c^2$ are also in A.P.

Concept understanding — Cosine Rule

Cosine Rule

Where it comes from

How it is used

Q.In △ABC\triangle ABC△ABC, if cot⁡A\cot AcotA, cot⁡B\cot BcotB, cot⁡C\cot CcotC are in A.P. then show that a2a^2a2, b2b^2b2, c2c^2c2 are also in A.P.

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Q.In $\triangle ABC$ , if $\cot A$ , $\cot B$ , $\cot C$ are in A.P. then show that $a^2$ , $b^2$ , $c^2$ are also in A.P.

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