If $\sum\limits_{n=1}^{2026}\tan^{-1}\left(\dfrac{1}{n^2+n+1}\right) = \tan^{-1}\left(1 - \dfrac{1}{x}\right)$, where $x \neq 0$, then $x = $

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

$\sin\left(3\sin^{-1}\left(\dfrac{1}{5}\right)\right) =$

$\cos^{-1}\left(\cos\dfrac{4\pi}{3}\right) + \sin^{-1}\left(\sin\dfrac{4\pi}{3}\right) = \ldots$