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)
$\endgroup$4 Answers
$\begingroup$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 $c_-=\pi$.
$\endgroup$ 1 $\begingroup$Hint. Let $t=\tan^{-1}(\sqrt{x^2-1})$ then $$x^2=1+\tan^2(t)=\frac{1}{\cos^2(t)}.$$
$\endgroup$ $\begingroup$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, $\sec^{-1}(x) = \theta = \tan^{-1}(\sqrt{x^2-1})$.
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 $\sec^{-1}(x) = \tan^{-1}\left(\sqrt{x^2-1}\right)$.
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