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

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Solution

sice $sin A . cos A$ and $tan A$ are G.P. we have ${cos}^{2} A = sin A tan A$
$\Rightarrow {cot}^{2} A = s e c A \Rightarrow {cot}^{4} A = 1 + {tan}^{2} A$
$\Rightarrow {cot}^{6} A - {cot}^{2} A = 1$

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