Trigonometry + Progressions = Trigonessions!

As they are in GP, $\dfrac{\cos\theta}{\sin\theta}=\dfrac{\tan\theta}{\cos\theta}\implies\cos^2\theta=\dfrac{\sin^2\theta}{\cos\theta}\\\boxed{\sin^2\theta=\cos^3\theta}$

We know, $\cos\theta$ is the Geometric mean of the other to values. $\therefore\sqrt{\sin\theta×\tan\theta}=\cos\theta\\\dfrac{\sin\theta}{\sqrt{\cos\theta}}=\cos\theta\\\tan\theta=\sqrt{\cos\theta}\implies\boxed{\cot^2\theta=\dfrac1{\cos\theta}}$

$\therefore\cot^6\theta-\cot^2\theta=\dfrac1{\cos^3\theta}-\dfrac1{\cos\theta}\\=\dfrac{\cos\theta(1-\cos^2\theta)}{\cos^4\theta}=\dfrac{\cos\theta×\sin^2\theta}{\cos^4\theta}\\=\dfrac{\cos\theta×\cos^3\theta}{\cos^4\theta}=\boxed1$

$\sin(\theta), \cos(\theta), \tan(\theta) = \sin(\theta), a * \sin(\theta), a^2 * \sin(\theta)$ for some $a(\theta)$ .

As $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = a^2 * \sin(\theta), a = \frac{1}{\sqrt{\cos(\theta)}}$ .

$\cos(\theta) = \frac{\sin(\theta)}{\sqrt{\cos(\theta)}}$ , so $\cos^3(\theta) = \sin^2(\theta)$ .

$\cot^6(\theta) - \cot^2(\theta) = \frac{(\cos^3(\theta))^2}{\sin^6(\theta)} - \frac{(\cos^2(\theta)}{\sin^2(\theta)}$

$= \frac{\sin^4(\theta)}{\sin^6(\theta)} - \frac{\cos^2(\theta)}{\sin^2(\theta)} = \frac{1 - \cos^2(\theta)}{\sin^2(\theta)} = 1$ .