If $\sin^{-1}(x - 2) + \cos^{-1}(x) + \tan^{-1}(x + 2) + \cot^{-1}(x + 4) = \sec^{-1}(\sqrt{k}) - \dfrac{\pi}{2}$, then $\cos(2\,\text{cosec}^{-1}\sqrt{k - 1}) = \ldots$

For the equation to be defined, the arguments of the inverse trigonometric functions must lie within their respective domains.

For $\sin^{-1}(x - 2)$, we require $-1 \le x - 2 \le 1 \Rightarrow 1 \le x \le 3$.

For $\cos^{-1}(x)$, we require $-1 \le x \le 1$.

The intersection of these two domains is $x = 1$.

Substituting $x = 1$ into the given equation:

$\sin^{-1}(1 - 2) + \cos^{-1}(1) + \tan^{-1}(1 + 2) + \cot^{-1}(1 + 4) = \sec^{-1}(\sqrt{k}) - \dfrac{\pi}{2}$

$\sin^{-1}(-1) + \cos^{-1}(1) + \tan^{-1}(3) + \cot^{-1}(5) = \sec^{-1}(\sqrt{k}) - \dfrac{\pi}{2}$

$-\dfrac{\pi}{2} + 0 + \tan^{-1}(3) + \tan^{-1}\left(\dfrac{1}{5}\right) = \sec^{-1}(\sqrt{k}) - \dfrac{\pi}{2}$

$\tan^{-1}(3) + \tan^{-1}\left(\dfrac{1}{5}\right) = \sec^{-1}(\sqrt{k})$

Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\dfrac{x + y}{1 - xy}\right)$ for $xy < 1$:

$\tan^{-1}\left(\dfrac{3 + \dfrac{1}{5}}{1 - 3 \times \dfrac{1}{5}}\right) = \sec^{-1}(\sqrt{k})$

$\tan^{-1}\left(\dfrac{16/5}{2/5}\right) = \sec^{-1}(\sqrt{k})$

$\tan^{-1}(8) = \sec^{-1}(\sqrt{k})$

Let $\theta = \tan^{-1}(8)$, then $\tan \theta = 8$.

$\sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + 64} = \sqrt{65}$.

Therefore, $\sec^{-1}(\sqrt{65}) = \sec^{-1}(\sqrt{k}) \Rightarrow k = 65$.

We need to find the value of $\cos(2\text{cosec}^{-1}\sqrt{k - 1})$:

$\cos(2\text{cosec}^{-1}\sqrt{65 - 1}) = \cos(2\text{cosec}^{-1}8)$

Let $\alpha = \text{cosec}^{-1}8$, then $\text{cosec} \alpha = 8 \Rightarrow \sin \alpha = \dfrac{1}{8}$.

Using the double angle formula $\cos(2\alpha) = 1 - 2\sin^2 \alpha$:

$\cos(2\alpha) = 1 - 2\left(\dfrac{1}{8}\right)^2 = 1 - 2\left(\dfrac{1}{64}\right) = 1 - \dfrac{1}{32} = \dfrac{31}{32}$.

Answer: $\dfrac{31}{32}$