# What is $\tan(\operatorname{arcsec} (x))$?

What is $\def\arcsec{\operatorname{arcsec}}\tan(\arcsec(x))$ simplified and why?

More specifically, I followed this reasoning, but apparently it is wrong:

$\tan(\arcsec(x))=\sqrt{\sec^2(\arcsec(x))-1}=\sqrt{x^2-1}$

What is wrong with this reasoning?

Apparently the answer is: $\sqrt{x^2-1}$ for $x\ge1$ and $-\sqrt{x^2-1}$ for $x\le1$ Why is this the right answer?

• it is $$\sqrt{1-\frac{1}{x^2}} x$$ – Dr. Sonnhard Graubner Feb 27 '16 at 16:20

By definition we have: $$y=\mbox{arcsec}x \iff \sec y=x \Rightarrow \cos y=\frac{1}{x}\Rightarrow \sin y=\sqrt{1-\frac{1}{x^2}} \Rightarrow$$
$$\Rightarrow \tan y = \frac{\sin y}{\cos y}=x\sqrt{1-\frac{1}{x^2}}=\frac{x}{|x|}\sqrt{x^2-1}$$
• You have forgot the possible two signs of the square root: $\tan \alpha=\pm \sqrt{\sec^2 \alpha-1}$, where the sign depend on the position of $\alpha$ – Emilio Novati Feb 27 '16 at 22:39
• But isn't always $sin(\alpha)=\sqrt{1-cos^2(\alpha)}$ without $\pm$? – GambitSquared Feb 27 '16 at 22:44
• What do you mean by the position of $\alpha$? – GambitSquared Feb 27 '16 at 22:45
• Hint: $\sin \alpha=\sqrt{1-\cos^2 \alpha}$ for $0\le\alpha \le \pi$, but change sign for $\pi<\alpha\le 2\pi$. – Emilio Novati Feb 27 '16 at 22:54
Try differentiating. Let $\def\arcsec{\operatorname{arcsec}}f(x)=\tan(\arcsec x)$. Then $$f'(x)=\sec^2(\arcsec x)\frac{1}{|x|\sqrt{x^2-1}}$$ (assuming $\arcsec$ is as the inverse of the secant over the set $[0,\pi/2)\cup(\pi/2,\pi]$), so $$f'(x)=\frac{|x|}{\sqrt{x^2-1}}$$ Therefore, for $x>1$, $$f(x)=\sqrt{x^2-1}+c_+$$ and, since $f(1)=\tan0=0$, we have $c_+=0$. For $x<-1$, $$f(x)=-\sqrt{x^2-1}+c_-$$ and, since $f(-1)=\tan\pi=0$, we have again $c_-=0$. Thus $$\tan(\arcsec x)=\operatorname{sgn}(x)\sqrt{x^2-1}$$