If a^2, b^2, c^2 are in A.P.,then which of th | vedclass

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(C) Given that $a^2, b^2, c^2$ are in $A.P.$,we have $2b^2 = a^2 + c^2$.Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substi

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$\cot A, \cot B, \cot C$

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(C) Given that $a^2, b^2, c^2$ are in $A.P.$,we have $2b^2 = a^2 + c^2$.Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these into the $A.P.$ condition: $2(2R \sin B)^2 = (2R \sin A)^2 + (2R \sin C)^2$.This simplifies to $2 \sin^2 B = \sin^2 A + \sin^2 C$,which means $\sin^2 B - \sin^2 A = \sin^2 C - \sin^2 B$.Using the identity $\sin^2 x - \sin^2 y = \sin(x+y)\sin(x-y)$,we get $\sin(B+A)\sin(B-A) = \sin(C+B)\sin(C-B)$.Since $A+B+C = \pi$,we have $\sin(B+A) = \sin C$ and $\sin(C+B) = \sin A$.Thus,$\sin C \sin(B-A) = \sin A \sin(C-B)$.Expanding this,$\sin C(\sin B \cos A - \cos B \sin A) = \sin A(\sin C \cos B - \cos C \sin B)$.Dividing both sides by $\sin A \sin B \sin C$,we obtain $\cot A - \cot B = \cot B - \cot C$.Therefore,$\cot A, \cot B, \cot C$ are in $A.P.$

If $a^2, b^2, c^2$ are in $A.P.$,then which of the following are also in $A.P.$?

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