Show that the gradient of the curve is positive for all x in the given interval (for a trig function) I am given the following question:

The equation of a curve is $\ y = \frac{\sin x}{1+\cos x}$ for $−π<x<π$. Show that
the gradient of the curve is positive for all $x$ in the given interval.

I've successfully differentiated the curve's equation to get:
$ \frac{dy}{dx} = \frac{1}{1 + cosx}$
My last issue is mathematically proving that the derivative function stays above 0 for all values of x. I understand that $\ -1 < \cos x < 1 $, so $\ 0 < \cos x + 1 < 2$. However, how do I figure out the max and min values that the derivative function $ \frac{dy}{dx} = \frac{1}{1 + cosx}$ can take?
P.S.: I'm still learning formatting equations and such into the correct format, and am using this place as an alternative to no teachers and peers due to the virus situation (have AP-equivalent calc exams soon).
 A: Firstly note that your function is undefined at $x=-\pi$. So I think the interval should be $-\pi<x<\pi$.
As @lab bhattacharjee mentioned in the comments we have $\cos(2x)=2\cos^{2}(x)-1$ so $$\cos(x)=2\cos^{2}(\frac{x}{2})-1$$ and $\cos(x)+1=2\cos^{2}(\frac{x}{2}).$
Clearly $\cos^{2}(\frac{\pi}{2})=\cos^{2}(\frac{-\pi}{2})=0$, $\cos^2(0)=1$ and it follows that $0<2\cos^2(\frac{x}{2})\leq2$ for all $x\in(-\pi,\pi).$
Hence $$\frac{dy}{dx}=\frac{1}{1+\cos(x)}=\frac{1}{2\cos^{2}(\frac{x}{2})}$$ and from the previous calculations we have $\frac{1}{2\cos^{2}(\frac{x}{2})}\geq\frac{1}{2}$ for $-\pi<x<\pi.$
A: We know that
$$\frac{\sin{x}}{1+\cos{x}}=\frac{2\sin\frac{x}{2} \cos\frac{x}{2}}{1+2\cos^2\frac{x}{2}-1}=\frac{2\sin\frac{x}{2} \cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=\tan\frac{x}{2}$$
And we know that
$$\frac{d}{dx}\tan\frac{x}{2}=\frac{1}{2}\sec^2\frac{x}{2}$$
As something squared is always positive, then the gradient must always be positive as required.
(Note: we have proved a stronger result; namely that the gradient is positive for all $x\in\mathbb R$, not just in the given interval.)
EDIT:
In answer to your comment:
We have
$$0<1+\cos{x}<2$$
Assuming you understand why this true, consider
$$\frac{1}{1+\cos{x}}$$
If $1+\cos{x}$ is very very very close to $2$, which is possible according to the inequality, then $\frac{1}{1+\cos{x}}$ is very very very close to- but very slightly greater- than $\frac{1}{2}$. (If you have trouble understanding why, here's another example: $\frac{1}{1.9}>\frac{1}{2}$.) So we have
$$\frac{1}{1+\cos{x}}>\frac{1}{2}$$
However, if $1+\cos{x}$ is very close to $0$, then the expression $\frac{1}{1+\cos{x}}$ gets extremely big; mathematically we say that as $1+\cos{x}$ approaches $0$, then $\frac{1}{1+\cos{x}}$ approaches infinity, or $\infty$ if you prefer.
Obviously, for the values of $1+\cos{x}$ in between $0$ and $2$, the value of $\frac{1}{1+\cos{x}}$ will be 'between' $\frac{1}{2}$ and $\infty$.
So, we can write that
$$\frac{1}{2}<\frac{1}{1+\cos{x}}<\infty$$
ie $$\frac{1}{2}<\frac{1}{1+\cos{x}}$$ as required. Remember; anything divide by $0$ is infinitely large, not $0$ (I believe that's the mistake that the examiners were alluding to). I hope that helps. If not, don't hesitate to ask any further questions.
Good luck in your A Levels, from one A-Level Student to another :)
A: Expressed  in terms of half angles
$$ y=\dfrac{2sc}{2cc}= \tan \frac x2$$
Its derivative $\dfrac{1}{2\cos^2(x/2)}$ is always $>0$ and increases monotonously between x-limits $\pm \pi. $
