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

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


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?

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

By definition we have: $$ y=\mbox{arcsec}x \iff \sec y=x \Rightarrow \cos y=\frac{1}{x}\Rightarrow \sin y=\sqrt{1-\frac{1}{x^2}} \Rightarrow $$

$$ \Rightarrow \tan y = \frac{\sin y}{\cos y}=x\sqrt{1-\frac{1}{x^2}}=\frac{x}{|x|}\sqrt{x^2-1} $$

  • $\begingroup$ Thanks! And what is wrong with my reasoning? $\endgroup$ – GambitSquared Feb 27 '16 at 22:22
  • $\begingroup$ 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 $\alpha$ $\endgroup$ – Emilio Novati Feb 27 '16 at 22:39
  • $\begingroup$ But isn't always $sin(\alpha)=\sqrt{1-cos^2(\alpha)}$ without $\pm$? $\endgroup$ – GambitSquared Feb 27 '16 at 22:44
  • $\begingroup$ What do you mean by the position of $\alpha$? $\endgroup$ – GambitSquared Feb 27 '16 at 22:45
  • $\begingroup$ Hint: $\sin \alpha=\sqrt{1-\cos^2 \alpha}$ for $0\le\alpha \le \pi$, but change sign for $\pi<\alpha\le 2\pi$. $\endgroup$ – 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 $$ \tan(\arcsec x)=\operatorname{sgn}(x)\sqrt{x^2-1} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.