When to apply supplementary/complementary angle identity for $sec^{-1}x$ So hopefully I can articulate my thoughts well enough for you to understand, if not feel free to ask and I will try clarify.
So I have the following problem where I have to show
$\sec^{-1}x= \tan^{-1}(\sqrt{x^2 -1})$ if $x\ge1$
and 
$\sec^{-1}x= \pi -\tan^{-1}(\sqrt{x^2 -1})$ if $x\le-1$
This might not have nothing to do with either angle identity, but I do not understand why the function when the condition $x\le-1$ is applied gets different. 
Also when should the supplementary/complementary angle identites be applied when solving trigonometric problems together with inverses? 
Take the number $\frac{4\pi}{5}$ as an example since the number does not reside between $[-\frac{\pi}{2},\frac{\pi}{2}]$ for $\sin^{-1}(\sin(\frac{4\pi}{5}))$ it could be redefined(?) as $\sin(\frac{4\pi}{5})=\sin(\pi-\frac{\pi}{5})=\sin(\frac{\pi}{5})$ and now it is usable. But why?
Many thanks whomever might help me!
 A: Let me assume your arcsecant function takes its values on $[0,\pi/2)\cup(\pi/2,\pi]$ (but see this answer for a different approach). In other words, for $x\ge1$, $\DeclareMathOperator\arcsec{arcsec}\arcsec x$ is the unique angle in $[0,\pi/2)$ such that $\sec(\arcsec x)=x$ and similarly for $x\le-1$.
Suppose $x\ge 1$ and set $\alpha=\arcsec x$; then
$$
\sec^2\alpha=1+\tan^2\alpha
$$
that is,
$$
\tan^2\alpha=\sec^2\alpha-1=x^2-1
$$
We can also say $\tan\alpha=\sqrt{x^2-1}$, because by assumption $0\le\alpha<\pi/2$, so
$$
\arcsec x=\alpha=\arctan\sqrt{x^2-1}
$$
Let's look at $x\le-1$; again, let $\alpha=\arcsec x\in(\pi/2,\pi]$. We can repeat the argument above until $\tan\alpha=-\sqrt{x^2-1}$, because the tangent is negative. Now $\arctan(-\sqrt{x^2-1})=-\arctan\sqrt{x^2-1}\in(-\pi/2,0]$, but we need to find the angle $\alpha\in(\pi/2,\pi]$ whose tangent is $-\sqrt{x^2-1}$, which is
$$
\pi-\arctan\sqrt{x^2-1}
$$
so we end up with the formula you needed to prove.
A: The basic problem here is that, strictly speaking, there is no such function as $\sin^{-1} x$ (and similar for the other trigonometric and many other functions). Strictly speaking, for a function $f:A\to B$ to have an inverse $f^{-1}:B\to A$ you need $f$ to be "one to one" (so that $f^{-1}$ is not ambiguous) and "onto" (so that $f^{-1}$ is defined on $B$ in the first place.
The trouble here is that $\sin:\mathbb R\to\mathbb R$ is neither "one to one" nor "onto".
To fix that, we introduce various conventions. For example, if we agree to restrict the domain and range of $\sin$ to $[-\frac{\pi}{2},\frac{\pi}{2}]$ and $[-1,1]$, respectively, that function (still conveniently but confusingly labelled $\sin$) is both "one to one" and "onto". Thus it has an inverse, and we call that inverse $\sin^{-1}$ or $\arcsin$. But remember, this new function is not the inverse of $\sin$. It is the inverse of "the other" $\sin$, which is the common $\sin$ with restricted domain and range.
Now it is worth studying how $\sin^{-1}$ and $\sin$ are related. You do know that $\sin^{-1}(\sin x)=x$ if $x\in[-\frac{\pi}{2},\frac{\pi}{2}]$ and you know that $\sin(\sin^{-1}x)=x$ for $x\in[-1,1]$. What you are asking is what the relationship between $\sin^{-1}(\sin x)$ and $x$ is, for arbitrary $x\in\mathbb R$. It can get complicated, and individual cases are handled exactly as in your example for $\frac{4\pi}{5}$. Because $\frac{4\pi}{5}\not\in[-\frac{\pi}{2},\frac{\pi}{2}]$, you cannot claim that $\sin^{-1}(\sin\frac{4\pi}{5})=\frac{4\pi}{5}$. However, you can find another number ($\frac{\pi}{5}$) which is in $[-\frac{\pi}{2},\frac{\pi}{2}]$ and has the same sine. (That is why the supplementary/complementary angle identities are useful for!) Thus, $\sin^{-1}(\sin\frac{4\pi}{5})=\sin^{-1}(\sin\frac{\pi}{5})=\frac{\pi}{5}$, the second equality following from the fact that $\sin^{-1}$ is the inverse of "the other" (restricted) $\sin$.
I believe, in your first example (with $\sec^{-1}$ and $\tan^{-1}$), the need to distinguish cases comes from the fact that, to define $\sec^{-1}$, by convention we restrict the domain to $[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$ and for $\tan^{-1}$ - to $(-\frac{\pi}{2},\frac{\pi}{2})$, i.e. to a different set.
