# Prove that $\tan (x)=\sin (x)+1$ has only one solution in $\left(−\frac π2,\frac π2\right)$

Prove that $\tan (x)= \sin (x) +1$ has only one solution in $\left(−\frac π2,\fracπ2\right)$ and we must use Rolle's rule and Cauchy's mean value theorem.

I know how to solve it without the use of Rolle's rule and Cauchy's mean value theorem. Can you help?

Let $f(x)=\tan(x)-\sin(x)-1$. You have $f'(x)=\sec^2(x)-\cos(x)$. Since $\sec^2(x)-\cos(x)$ has the same sign as $x$ for $x\in[\frac{-\pi}{2},\frac{\pi}{2}]$, we know that $f'(x)\geq0$ in this interval and $f'(x)>0$ for $x\in[\frac{-\pi}{2},\frac{\pi}{2}]\setminus\{0\}$. Therefore $f$ is strictly increasing and can have at most one zero.
Using Rolle's Theorem, it does not satisfy the hypothesis that $f$ is differnetiable for all vales of $x$ in the open interval $(a,b)$, $f$ is continuous at $x=a$ or $x=b$, and that $f(a)=f(b)=0$; however the conclusion can be true even though the hypothesis is false for $f$ stating that there exists one number $x=c$ in $(a,b)$ such that $f'(c)=0$.
Using the Mean Value Theorem, it does not satisfy the hypothesis $f$ is differnetiable for all vales of $x$ in the open interval $(a,b)$, $f$ is continuous at $x=a$ or $x=b$; however satisfies the conclusion that there exists at least one number $x=c$ in $(a,b)$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a}.$$
Hint: $f(x) = \tan x - \sin x - 1$ has $f'(x) = \sec^2 x - \cos x=\dfrac{1-\cos^3 x}{\cos^2 x} \geq 0$, and $f(-\pi/4) = -2 +\dfrac{1}{\sqrt{2}} < 0$, and since $\lim_{ x \to \pi/2^{-}} f(x) = + \infty$, we have for $M = 1$, you can find a delta $\delta > 0$ such that $0 < |x- \pi/2| < \delta \Rightarrow f(x) > M = 1$. Thus by continuity of $f$ and by using MVT on $(-\pi/4, \pi/2-\epsilon)$, there is a root and coupled with $f'(x) > 0$, there is only one root, with $\epsilon$ can be selected easily from $\delta$.