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If $\sin A, \cos A$ and $\tan A$ are in G.P., then
$\cot^{6} A - \cot^{2} A =$

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Correct option is C

cos2⁡A=sin⁡Atan⁡A=sin2⁡A⋅sec⁡A⇒cot2⁡A=sec⁡A∴ cot6⁡A−cot2⁡A=cot2⁡Acot4⁡A−1 =cot2⁡Asec2⁡A−1 =cot2⁡A×tan2⁡A=1

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If sin⁡A,cos⁡A and tan⁡A are in G.P., then cot6⁡A−cot2⁡A=

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If $\sin A, \cos A$ and $\tan A$ are in G.P., then
$\cot^{6} A - \cot^{2} A =$