If $$\cos A, \sin A,\cot A$$ are in geometric progression, then the value of $$\tan^6 A - \tan^2 A$$ is:

Given,  $$\cos A, \sin A,\cot A$$ are in geometric progression

$$=$$>  $$\sin^2A=\cos A\cot A$$

$$=$$>  $$\sin^2A=\cos A\frac{\cos A}{\sin A}$$

$$=$$>  $$\sin^3A=\cos^2A$$ .....................(1)

$$\tan^6A-\tan^2A=\frac{\sin^6A}{\cos^6A}-\frac{\sin^2A}{\cos^2A}$$

$$=\frac{\left(\sin^3A\right)^2}{\cos^6A}-\frac{1-\cos^2A}{\cos^2A}$$

$$=\frac{\left(\cos^2A\right)^2}{\cos^6A}-\frac{1}{\cos^2A}+1$$

$$=\frac{\cos^4A}{\cos^6A}-\frac{1}{\cos^2A}+1$$

$$=\frac{1}{\cos^2A}-\frac{1}{\cos^2A}+1$$

$$=1$$

Hence, the correct answer is Option A