Prove with the Mean Value Theorem that the function $\tan(x)$ increases in the interval $(\frac{-\pi}{2}, \frac{\pi}{2})$.

My problem is that to use the Mean Value Theorem you need $f(\frac{\pi}{2})$ and $f(\frac{-\pi}{2})$ but in those values it's undefined. I asked if I could use a smaller interval but I was told I must to use $(\frac{-\pi}{2}, \frac{\pi}{2})$.

Thanks for reading.


We have $\tan(x) = \frac{\sin(x)}{\cos(x)}$, hence $\tan'(x) = \frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)} = \frac 1{\cos^2(x)}$. Let $x,y\in (-\pi/2,\pi/2)$, $x<y$. Then there is some $\xi\in (x,y)$ such that $$ \tan(y)-\tan(x) = \tan'(\xi)(y-x) = \frac{y-x}{\cos^2(\xi)}> 0. $$ Hence, $\tan(x)<\tan(y)$.

  • $\begingroup$ I have a doubt, does the mean value theorem really apply on this interval? in my book, the hypothesis states that the function has to be continuous on the interval $[a,b]$ and differentiable on $(a,b)$, but $tanx$ is not continuous on $[\frac{-\pi}{2},\frac{\pi}{2}]$ because $\lim_{x\to {-\pi/2}^{+}} tanx = -\infty$ and $\lim_{x\to {\pi/2}^{-}} tanx = \infty$. so i'm not sure whether it's legal to apply the mean value theorem there. $\endgroup$ – Donlans Donlans Oct 8 '19 at 1:10
  • 1
    $\begingroup$ $f(x) = \tan x$ is continuous and differentiable on $(-\frac{\pi}{2},\frac{\pi}{2})$. So if $x,y \in (-\frac{\pi}{2},\frac{\pi}{2})$ and $x<y$, then $f(x) = \tan x$ is continuous on $[x,y]$ and differentiable on $(x,y)$. $\endgroup$ – JDZ Oct 8 '19 at 2:27

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.