The area of the triangle formed by the lines $y=x, y=0\text{ and }x={\sin }^{-1}\left(a^4+1\right)+{\cos }^{-1}\left(a^4+1\right)-{\tan }^{-1}\left(a^4+1\right)$ is sq. units, where$k$ =

We observe that  $a^4+1\ge 1$ for all $a\in R$ Therefore $x={\sin }^{-1}\left(a^4+1\right)+{\cos }^{-1}\left(a^4+1\right)-{\tan }^{-1}\left(a^4+1\right)$ is meaning it for $a= 0$and in that case it becomes $y=\frac{\pi }{4}$. Thus, the given lines $y=x,y=0\text{ and }x=\frac{\pi }{4}$ These lines form a right triangle area $\frac{{\pi }^2}{32}$ sq. umts