If  $\alpha$ satisfies the inequation  $x^2-x-2>0$, then a value exists for

$x^2-x-2>0$

$\Rightarrow (x-2)(x+1)>0$

$\Rightarrow x<-1\text{\hspace{0.17em}}orx>2$, using sign scheme

We know that  ${\sin }^{-1}x$ is defined for  $\left|x\right|\le 1,{\cos }^{-1}x$ is define for

$\left|x\right|\le 1and{\sec }^{-1}x$ $\left|x\right|\ge 1,i.e.,x\le -1orx\ge 1$

Hence, (B) is the correct answer