why does $\tan^{-1}(\sqrt{x^2-1})=\sec^{-1}(x)$ 
Why does $\tan^{-1}(\sqrt{x^2-1})=\sec^{-1}(x)$

How do you derive $\tan^{-1}(\sqrt{x^2-1})$ from $\sec^{-1}(x)$ ? 
(By $^{-1}$ I mean the inverse function)
 A: Hint. Let $t=\tan^{-1}(\sqrt{x^2-1})$ then 
$$x^2=1+\tan^2(t)=\frac{1}{\cos^2(t)}.$$
A: I find this type of problem easiest to understand if I draw a figure.  So suppose $\theta = \tan^{-1}(\sqrt{x^2-1})$, i.e. $\tan(\theta) = \sqrt{x^2-1}$.  That gives us a right triangle with height $\sqrt{x^2-1}$ and base $1$.  By the Theorem of Pythagorus, the hypotenuse has length $\sqrt{(x^2-1)+1}=x$.  So from the figure, $\sec(\theta) = x/1$, or equivalently, $\sec^{-1}(x) = \theta = \tan^{-1}(\sqrt{x^2-1})$.

A: If you define the arcsecant to have values in $[0,\pi/2)\cup(\pi/2,\pi]$, then the identity you write can only be true for $x\ge1$, but definitely not for $x\le-1$.
Consider $f(x)=\operatorname{arcsec}x$ defined for $|x|\ge1$ and with values as stated above. Then $\sec f(x)=x$, that is,
$$
\cos f(x)=\frac{1}{x}
$$
and therefore
$$
f'(x)\sin f(x)=\frac{1}{x^2}
$$
Now $0\le f(x)\le\pi$ and $\sin f(x)=\sqrt{1-\cos^2f(x)}$, so
$$
\sin f(x)=\sqrt{1-\frac{1}{x^2}}=\frac{\sqrt{x^2-1}}{|x|}
$$
which means that
$$
f'(x)=\frac{1}{|x|\sqrt{x^2-1}}
$$
With easy calculations, if $g(x)=\arctan\sqrt{x^2-1}$, we find
$$
g'(x)=\frac{1}{x\sqrt{x^2-1}}
$$
Thus
$$
f'(x)=\begin{cases}
g'(x) & x>1 \\
-g'(x) & x<-1
\end{cases}
$$
which means there are constants $c_+$ and $c_-$ such that
$$
\operatorname{arcsec}x=\begin{cases}
c_+ + \arctan\sqrt{x^2-1} & x\ge1 \\[4px]
c_- - \arctan\sqrt{x^2-1} & x\le-1
\end{cases}
$$
We can determine the constants by evaluating at $1$ and $-1$, so $c_+=0$ and $c_-=\pi$.
A: Let $\sec^{-1}(x) = y$, so $\sec(y) = x$. Using the Pythagorean Identity $\color{blue}{\sec^2(x) = \tan^2(x)+1}$, this can be rewritten as $\pm\sqrt{\tan^2(y)+1} = x$. Manipulating the equality yields $\tan(y) = \sqrt{x^2-1} \iff y = \tan^{-1}\left(\sqrt{x^2-1}\right)$, which means $\sec^{-1}(x) = \tan^{-1}\left(\sqrt{x^2-1}\right)$.
