Increasing nature of $\sec^{-1}x$ In most of the textbooks, it is said that $\sec^{-1}x$ is a increasing function.
But my doubt is that $\sec^{-1}(-1)=\pi$ and $\sec^{-1}{1}=0$ which is violating the fundamental property of increasing function 
$$f(x_1)<f(x_2)$$
$$x_1<x_2$$
 A: For clarification: I assume with $\sec^{-1}$ you mean the inverse function.
Then let's study $\sec$ itself first: $\sec$ itself is is periodic with periode $2\pi$, having poles at $n\pi + \frac{\pi}{2}$
Due to the periodicity the $\sec$-function itself is not injective hence to consider an inverse function we have to restrict the definition area and usually the intervall $[0,\pi]\setminus{\left\{\frac{\pi}{2}\right\}}$ is considered. 
On this intervall it holds: $$\sec: [0,\pi]\setminus{\left\{\frac{\pi}{2}\right\}} \to \Bbb{R}\setminus{(-1,1)}$$ but if we take a closer look to the part before and after the pole we get:
$$\sec: [0,\frac{\pi}{2}) \to [1,\infty)$$ and  $$\sec: (\frac{\pi}{2},\pi] \to (-\infty,-1]$$
What we see: The $\sec$ function has jumps at the poles. 
Hence for the inverse it holds: 
$$\sec^{-1}:  [1,\infty)\to [0,\frac{\pi}{2})$$ and  $$\sec^{-1}: (-\infty,-1] \to (\frac{\pi}{2},\pi]$$
With a jumps as well!
So while $\sec^{-1}(-1) = \pi$ the other part starts with $\sec^{-1}(1) = 0$
And we realize: Although it holds altogether that $$\sec^{-1}: \Bbb{R}\setminus{(-1,1)} \to [0,\pi]\setminus{\left\{\frac{\pi}{2}\right\}} $$ and $\sec^{-1}$ is strictly increasing on each of its definition areas, it jumps when crossing the non-defined area $(-1,1)$
