Prove that $\tan x =\sin x + 1$ have only one solution in $(-\frac{\pi}{2},\frac{\pi}{2})$ Prove that $\tan x = \sin x + 1$ have only one solution in $(-\frac{\pi}{2},\frac{\pi}{2})$

Well, I believe I need to show that there's a solution. so I need to find x which solves $\tan x = \sin x + 1$.

Then, I need to derivate it and show that the derivation either bigger than zero, or smaller than zero, Thus - there will be no solutions.

So in intuition way, I know how to solve it, but practically I'm having difficulties. Can you guys help with the steps I've shown?

I believe It's Rolle's rule. Because we just studied it :)
let me rewrite the equation as $$\tan t = \sin t + 1.\tag 1$$  i will use $x = \cos t, y = \sin t.$ then $(1)$ becomes $$\frac y x = y + 1 \to y = \frac x{1-x}\tag 2 $$

the solution to $(1)$ are the points where the hyperbola (2) cuts the unit circle in the first quadrant. this is easy to see because the  branch of the hyperbola going through $(0,0)$ can only cut the unit circle at one in the first quadrant and the second time at a pint in the third quadrant.
$\tan(x)$ is a bijection from $]-\pi/2, \pi/2[$ to $]-\infty, \infty[$ as a continous and increasing function (derivative >0). $\sin(x)+1$ is increasing, continuous function: this is a bijection from $]-\pi/2, \pi/2[$ to $]0,2[$.

hence, the two graphs must have a unique common point as $\sin(x)+1 \in [0, 2|$. To have an interval, you apply the Bolzano's theorem:

if f is continuous on [a,b] and f(a)&lt; 0 and f(b)>0 then, there is $\alpha$ where $f(\alpha)=0$. $\alpha$ is unique if f is a bijection on [a,b] (example : f monotone).

here $f(x)=\tan(x)-\sin(x)-1$. we have $\alpha=1.08264952...$