VerifiedHint: Use the basic definition of GP and the necessary trigonometric identities to arrive at the answer.
Complete step-by-step answer:
Now, we know that if 3 terms a, b, c are in GP then we can write,
$\dfrac{b}{a} = \dfrac{c}{b} = r$, where r is the common ratio.
The above can also be written as
$\Rightarrow b^2 = ac \to (1)$
It is given that $\sin \theta$, $\cos \theta$, $\tan \theta$ are in GP, so from eqn (1), we can write,
$\cos^{2} \theta = \sin \theta \times \tan \theta \to (2)$
Now $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$, substituting in equation (2), we get,
$\Rightarrow \cos^{2} \theta = \sin \theta \times {\dfrac{\sin \theta}{\cos \theta}}$
$\Rightarrow \cos^{2} \theta = \dfrac{\sin^{2} \theta}{\cos \theta}$
$\Rightarrow \dfrac{\sin^{2} \theta}{\cos^{3} \theta} =1 \to (3)$
It can be also rearranged as,
$\Rightarrow \dfrac{\cos^{2} \theta}{\sin^{2} \theta} = \dfrac{1}{\cos \theta} = \sec \theta$
$\Rightarrow \cot^{2} \theta = \sec \theta \to (4) (\because \dfrac{\cos \theta}{\sin \theta} = \cot \theta)$
Now, we have to find the value of $\cot^{6} \theta - \cot^{2} \theta$
It can be written as, $(\cot^{2} \theta)^3 - \cot^{2} \theta$
Substituting from eqn (4), we get,
$\Rightarrow \sec^{3} \theta - \sec \theta$
Taking $\sec \theta$ common
$\Rightarrow \sec \theta (\sec^{2} \theta - 1) $
We know (from trigonometric identities) that $\sec^{2} \theta - 1 = \tan^{2} \theta$, we get
$\Rightarrow \sec \theta \times \tan^{2} \theta $
$\Rightarrow \dfrac{1}{\cos \theta} \times \dfrac{\sin^{2} \theta}{\cos^{2} \theta}$
$\Rightarrow \dfrac{\sin^{2} \theta}{\cos^{3} \theta}$
$= 1$ (from eqn. (3))
Hence the value of $\cot^{6} \theta - \cot^{2} \theta$ is 1.
$\therefore$ Option A. is correct.
Note: Such problems, where more than one concept is involved can be solved easily by knowing the basics of those concepts. Using the necessary properties will solve these problems. Mistakes can be avoided while rearranging and substituting.
