In Delta ABC, if cot A, cot B, cot C are in A | vedclass

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(C) Given that $\cot A, \cot B, \cot C$ are in $A.P.$This implies $\cot A + \cot C = 2\cot B.$Using the identity $\cot A = \frac{\cos A}{\sin A}$ and

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$A.P.$

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(C) Given that $\cot A, \cot B, \cot C$ are in $A.P.$This implies $\cot A + \cot C = 2\cot B.$Using the identity $\cot A = \frac{\cos A}{\sin A}$ and the Sine Rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$We have $\cot A = \frac{b^2 + c^2 - a^2}{4\Delta}$ and $\cot B = \frac{a^2 + c^2 - b^2}{4\Delta}$ and $\cot C = \frac{a^2 + b^2 - c^2}{4\Delta},$where $\Delta$ is the area of the triangle.Substituting these into the $A.P.$ condition:$\frac{b^2 + c^2 - a^2}{4\Delta} + \frac{a^2 + b^2 - c^2}{4\Delta} = 2 \left( \frac{a^2 + c^2 - b^2}{4\Delta} \right)$$b^2 + c^2 - a^2 + a^2 + b^2 - c^2 = 2(a^2 + c^2 - b^2)$$2b^2 = 2(a^2 + c^2 - b^2)$$b^2 = a^2 + c^2 - b^2$$2b^2 = a^2 + c^2$This condition implies that $a^2, b^2, c^2$ are in $A.P.$

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

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