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So there's this question that asks us to prove that, between any two roots of $\tan x=1$ there exists at least one root of $\tan x =-1$. Suppose we assume that $a,b$ are two roots of $\tan x-1=0$, then $f(a)=f(b)=0$, where $f(x)= \tan x-1$. According to the theorem, $f'(c)=0$ where $c \in (a,b)$,i.e., $\sec^2 c =0$....and this is not defined. Either there's something wrong with my understanding or with the problem. Please help.

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    $\begingroup$ The function is not continuous on some intervals. It has infinite discontinuities at $x=\frac{k\pi}{2}$ for some $k\in\mathbb{Z}-\{0\}$. So Rolle's theorem does not apply. $\endgroup$ – C Squared Aug 28 '20 at 8:31
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The function $f(x)=\tan(x)-1$ is not continuous for every $x\in[a,b]$, ($a,b$ being consecutive roots of $f$), so Rolle's Theorem cannot be applied in such interval.

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Let $f(x)=\tan x -1$, and $g(x)=\tan x +1$.

The roots of $f(x)$ occur in the interval $I=[\frac{\pi}{4}+\pi k, \frac{\pi}{2}+\pi k)\bigcup (\frac{\pi}{2}+\pi k, \frac{5\pi}{4}+\pi k]$ for some $k\in\mathbb{Z}$. The roots of $f(x)$ occur at the endpoints of interval $I$.

The roots of $g(x)$ occur in the interval $J=[-\frac{\pi}{4}+\pi k, \frac{\pi}{2}+\pi k)\bigcup(\frac{\pi}{2}+\pi k, \frac{3\pi}{4}+\pi k]$ for some $k\in\mathbb{Z}$. The roots of $g(x)$ occur at the endpoints of interval $J$.

The interval $(\frac{\pi}{2}+\pi k, \frac{3\pi}{4}+\pi k]$ is contained in $[\frac{\pi}{4}+\pi k, \frac{\pi}{2}+\pi k)\bigcup (\frac{\pi}{2}+\pi k, \frac{5\pi}{4}+\pi k]$ and a root of $g(x)$ occurs at $x=\frac{3\pi}{4}+\pi k$. Roots of $f(x)$ occur at $x=\frac{\pi}{4}+\pi k$ and $x=\frac{5\pi}{4}+\pi k$.

$\frac{\pi}{4}+\pi k<\frac{3\pi}{4}+\pi k<\frac{5\pi}{4}+\pi k $.

So there is at least one root of $g(x)$ between any two roots of $f(x)$.

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