Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$.

If $\sec (\theta+\alpha), \sec \theta$ and $\sec (\theta-\alpha)$ are in arithmetic progression,then $\sin ^2 \theta=$

For $0 < x \leq \pi$, $\sinh ^{-1}(\cot x)$ is equal to

If $\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha$ and $\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha$,then $\theta$ is equal to

For any positive integer $n$,let $S_n: (0, \infty) \rightarrow R$ be defined by $S_n(x) = \sum_{k=1}^n \cot^{-1}\left(\frac{1+k(k+1)x^2}{x}\right)$,where for any $x \in R$,$\cot^{-1} x \in (0, \pi)$ and $\tan^{-1} x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then which of the following statements is (are) $TRUE$?
$(A)$ $S_{10}(x) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1+11x^2}{10x}\right)$,for all $x > 0$
$(B)$ $\lim_{n \rightarrow \infty} \cot(S_n(x)) = x$,for all $x > 0$
$(C)$ The equation $S_3(x) = \frac{\pi}{4}$ has a root in $(0, \infty)$
$(D)$ $\tan(S_n(x)) \leq \frac{1}{2}$,for all $n \geq 1$ and $x > 0$

If $\sin \theta + \text{cosec} \theta = 2$,the value of $\sin^{10} \theta + \text{cosec}^{10} \theta$ is