If $y=\sec^{-1}\left(\frac{x+1}{x-1}\right)+\sin^{-1}\left(\frac{x-1}{x+1}\right)$, $x\in[0,\infty)$ and $x\ne1$, then $\dfrac{dy}{dx}$ is equal to

Recall the identity:
$\sec^{-1}(\theta) = \cos^{-1}\left(\frac{1}{\theta}\right)$
Applying this:
$\sec^{-1}\left(\frac{x+1}{x-1}\right) = \cos^{-1}\left(\frac{x-1}{x+1}\right)$

Substituting back:
$y = \cos^{-1}\left(\frac{x-1}{x+1}\right) + \sin^{-1}\left(\frac{x-1}{x+1}\right)$

Using the standard identity:
$\sin^{-1}(\theta) + \cos^{-1}(\theta) = \dfrac{\pi}{2}$

Therefore:
$y = \dfrac{\pi}{2}$

Differentiating:

$\dfrac{dy}{dx} = \dfrac{d}{dx}\left(\dfrac{\pi}{2}\right)$

$\boxed{\dfrac{dy}{dx} = 0}$