Question

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

Answer 1

Let $a=0$
We get
$x=si{n}^{-1}\left(1\right)co{s}^{-1}\left(1\right)-ta{n}^{-1}\left(1\right)$

$=\frac{\pi }{2}.0-\frac{\pi }{4}$
$=\frac{-\pi }{4}$
Hence
$sec\left(x\right)=\sqrt{2}$
$\Rightarrow se{c}^2\left(x\right)=2$