Cosinecos

$A=\cos^{-1} \left( \sin \left( \cos^{-1}x \right) \right) + \sin^{-1} \left( \cos \left( \sin^{-1}x \right) \right)$

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

Similarily

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

Hence adding both we get

$A=\pi -sin^{-1}x-cos^{-1}x$

also $sin^{-1}x+cos^{-1}x=\dfrac{\pi}{2}$

$A=\dfrac{\pi}{2}$

$tan(\dfrac{A}{2})=1$

Note- $\sin^{-1}x \in \left[ 0, \frac{\pi}{2} \right]$ is an important condition for 2nd and 3rd step above

Let $\begin{cases} \sin^{-1} x = \alpha & \Rightarrow \sin \alpha = x \\ \cos^{-1} x = \beta & \Rightarrow \cos \beta = x \end{cases}$

$\begin{aligned} \Rightarrow \sin \alpha & = \cos \beta = x \\ \cos \left(\frac{\pi}{2}-\alpha \right) & = \cos \beta \\ \Rightarrow \frac{\pi}{2}-\alpha & = \beta \\ \alpha + \beta & = \frac{\pi}{2} \end{aligned}$

$\begin{aligned} y & = \tan \left[ \frac{\cos^{-1} \left(\sin (\cos^{-1} x) \right) + \sin^{-1} \left(\cos (\sin^{-1} x) \right)}{2} \right] \\ & = \tan \left[ \frac{\cos^{-1} \left(\sin \beta \right) + \sin^{-1} \left(\cos \alpha \right)}{2} \right] \\ & = \tan \left[ \frac{\cos^{-1} \left(\cos \left(\frac{\pi}{2} - \beta \right) \right) + \sin^{-1} \left(\sin \left( \frac{\pi}{2} - \alpha \right)\right)}{2} \right] \\ & = \tan \left[ \frac{\cos^{-1} \left(\cos \alpha \right) + \sin^{-1} \left(\sin \beta \right)}{2} \right] \\ & = \tan \left[ \frac{\alpha + \beta}{2} \right] \\ & = \tan \frac{\pi}{4} = \boxed{1} \end{aligned}$