Q68 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics

In $\triangle ABC$, with usual notation, if cot A, cot B, cot C are in arithmetic progression, then

  • A
    sin A, sin B, sin C are in arithmetic progression.
  • Correct answer
    B
    $a^2, b^2, c^2$ are in arithmetic progression.
  • C
    cos A, cos B, cos C are in arithmetic progression.
  • D
    a, b, c are in arithmetic progression.

Solution

Given $\cot A, \cot B, \cot C$ are in A.P.

We know that $\cot A = \dfrac{\cos A}{\sin A}$.

Using the cosine rule, $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$ and using the sine rule, $\sin A = \dfrac{a}{2R}$.

Therefore, $\cot A = \dfrac{b^2 + c^2 - a^2}{2bc} \times \dfrac{2R}{a} = \dfrac{R(b^2 + c^2 - a^2)}{abc}$.

Similarly, $\cot B = \dfrac{R(c^2 + a^2 - b^2)}{abc}$ and $\cot C = \dfrac{R(a^2 + b^2 - c^2)}{abc}$.

Since $\cot A, \cot B, \cot C$ are in A.P., we have:

$2 \cot B = \cot A + \cot C$

Substituting the expressions, we get:

$2 \left( \dfrac{R(c^2 + a^2 - b^2)}{abc} \right) = \dfrac{R(b^2 + c^2 - a^2)}{abc} + \dfrac{R(a^2 + b^2 - c^2)}{abc}$

Cancelling the common term $\dfrac{R}{abc}$ from both sides:

$2(c^2 + a^2 - b^2) = (b^2 + c^2 - a^2) + (a^2 + b^2 - c^2)$

$2c^2 + 2a^2 - 2b^2 = 2b^2$

$2a^2 + 2c^2 = 4b^2$

$a^2 + c^2 = 2b^2$

This implies that $a^2, b^2, c^2$ are in arithmetic progression.

Answer: $a^2, b^2, c^2$ are in arithmetic progression.

Hence, the correct answer is option: 'B'.

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