In triangle ABC, with usual notation, if cot | vedclass

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(B) Given that $\cot A, \cot B, \cot C$ are in arithmetic progression, we have $2 \cot B = \cot A + \cot C$.Using the identity $\cot 	heta = \frac{\c

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

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(B) Given that $\cot A, \cot B, \cot C$ are in arithmetic progression, we have $2 \cot B = \cot A + \cot C$.Using the identity $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$, we get $2 \frac{\cos B}{\sin B} = \frac{\cos A}{\sin A} + \frac{\cos C}{\sin C} = \frac{\sin C \cos A + \cos C \sin A}{\sin A \sin C} = \frac{\sin(A+C)}{\sin A \sin C}$.Since $A+B+C = \pi$, $\sin(A+C) = \sin(\pi - B) = \sin B$.Thus, $2 \frac{\cos B}{\sin B} = \frac{\sin B}{\sin A \sin C}$, which implies $2 \cos B \sin A \sin C = \sin^2 B$.Using $2 \sin A \sin C = \cos(A-C) - \cos(A+C) = \cos(A-C) + \cos B$, we have $\cos B(\cos(A-C) + \cos B) = \sin^2 B = 1 - \cos^2 B$.$\cos B \cos(A-C) + \cos^2 B = 1 - \cos^2 B \implies \cos B \cos(A-C) = 1 - 2 \cos^2 B = -\cos(2B) = -\cos(2\pi - 2(A+C)) = \cos(2(A+C))$.Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, we know $\cot A = \frac{b^2+c^2-a^2}{4\Delta}$.Substituting this into the $AP$ condition: $\frac{b^2+c^2-a^2}{4\Delta} + \frac{a^2+b^2-c^2}{4\Delta} = 2 \frac{a^2+c^2-b^2}{4\Delta}$.$2b^2 = a^2+c^2$, which means $a^2, b^2, c^2$ are in arithmetic progression.

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

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