If $\sin A,\cos A,\tan A$ are in $G.P$  then ${\cot }^{6}A-{\cot }^{2}A=$ 1

$\sin A,\cos A,\tan A$ are in $G.P$

${\cot }^{6}A-{\cot }^{2}A$

${\cos }^{2}A=\sin A.\tan A\left(G.P\right)$

$=\sin A.\frac{\sin A}{\cos A}$

${\cos }^{3}A={\sin }^{2}A\Rightarrow \frac{{\cos }^{3}A}{{\sin }^{2}A}=1\to \left(1\right)$

$\frac{{\cos }^{2}A}{{\sin }^{2}A}.\cos A=1$

${\cot }^{2}A.\cos A=1\Rightarrow {\cot }^{2}A=\frac{1}{\cos A}=\sec A$

${\cot }^{6}A-{\cot }^{2}A={\left({\cot }^{2}A\right)}^3-{\cot }^{2}A$

$={\left(\sec A\right)}^3-\sec A$

$=\sec A\left({\sec }^{2}A-1\right)$

$=\sec A.{\tan }^{2}A$

$=\frac{1}{\cos A}.\frac{{\sin }^{2}A}{{\cos }^{2}A}=\frac{{\sin }^{2}A}{{\cos }^{3}A}=1\left(\because \text{Eqn}\text{.}\left(1\right)\right)$