If $x={\sin }^{-1}\left(a^6+1\right)+{\cos }^{-1}\left(a^4+1\right)-{\tan }^{-1}\left(a^2+1\right),a\in R$ then the value of ${\sec }^{2}x=$

$$\begin{gathered} Since {\sin }^{-1}is defined on \left[-1,1\right], we must have -1\le a^6+1\le 1 \\ \Rightarrow -2\le a^6\le 0 \\ \Rightarrow a^6=0 \\ \Rightarrow a=0 \\ \therefore x={\sin }^1\left(1\right)+{\cos }^{-1}\left(1\right)-{\tan }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{2}+0-\frac{\mathrm{\pi }}{4} \\ =\frac{\mathrm{\pi }}{4} \\ \therefore se{c}^{2}x=se{c}^2\left(\frac{\mathrm{\pi }}{4}\right)=2 \end{gathered}$$