If $\mathrm{x}={\sin }^{-1}({\mathrm{a}}^6+1)+{\cos }^{-1}({\mathrm{a}}^4+1)-{\tan }^{-1}({\mathrm{a}}^2+1),\mathrm{a}\in \mathrm{R}$ then the value of sec² x is equal to 

 Since sin^-1 is defined for [- 1, 1], so we have

a=0

$\begin{array}{rl} & \therefore \mathrm{x}={\sin }^{-1}1+{\cos }^{-1}1-{\tan }^{-1}1=\frac{\mathrm{\pi }}{4}\\ & \Rightarrow {\sec }^2\mathrm{x}=2\end{array}$