If $(\tan^{-1}x)^2 + (\cot^{-1}x)^2 = \dfrac{5\pi^2}{8}$, then the value of $x$ is equal to...

$\cos\left(\cos^{-1}\left(-\dfrac{1}{2}\right) + \dfrac{\pi}{3}\right) =$

$\sin^{-1}\left(\dfrac{12}{13}\right) + \cos^{-1}\left(\dfrac{4}{5}\right) + \tan^{-1}\left(\dfrac{63}{16}\right) =$

The value of $\tan^{-1}\left(\dfrac{\cos\left(\frac{19\pi}{4}\right) - 1}{\sin\left(\frac{\pi}{4}\right)}\right)$ is equal to