Trigonometric inequality $\frac{\cos x -\tan^2(x/2)}{e^{1/(1+\cos x)}}>0$ How can I solve the following inequality?

$$\frac{\cos x -\tan^2(x/2)}{e^{1/(1+\cos x)}}>0$$

 A: I'll post what I have so far so that it may give you an idea of what to do. It's not complete and I'll try and finish it of when I have more time.
$e^{1/(1+\cos x)}$ will always be positive, so we only have to worry about the numerator.
Using the Weierstrass substitution, as @Lucian suggested, $t=\tan{\frac{x}{2}}$ and keeping in mind that $\displaystyle \cos x = \frac{1-t^2}{1+t^2}$ we get:
$\displaystyle {\cos x -\tan^2(x/2)} = {\frac{1-t^2}{1+t^2} - t^2}$
$\displaystyle \frac{1-t^2}{1+t^2} - t^2 = \frac{1-t^2-t^2(1+t^2)}{1+t^2}$
So the fraction is going to be negative only when $1-t^2-t^2(1+t^2)$ is negative.
$1-t^2-t^2(1+t^2) \lt 0$, I´ll do another substitution to ease the calculations:
$u=t^2, 1-u-u(1+u)= -u^2-2u+1$
The roots are $u=-1 \pm \sqrt2 $ and because $-u^2-2u+1$ is concave downwards so it´s positive in $[-1 - \sqrt2,-1 + \sqrt2]$
Substituting back, and I'm assuming you are working in $\mathbb R, t=\pm \sqrt{-1 + \sqrt2}$
So $\displaystyle \frac{1-t^2}{1+t^2} - t^2 \gt 0, \  \ \forall t \in \left[-\sqrt{-1 + \sqrt2},\sqrt{-1 + \sqrt2} \right] $
A: Note that 
$$\frac{\cos x -\tan^2(x/2)}{e^{1/(1+\cos x)}} =\frac{ (-1 + 2 \cos x+ \cos^2 x)}{(1 + \cos x)} \cdot e^{-1/(1+\cos x)}$$
$\frac{e^{-1/(1+\cos x)}}{(1 + \cos x)} $ should be taken as a whole, because it is defined everywhere. It is $>0$ everywhere except at $(2k+1)\pi$ where it's zero. Therefore, we can focus  on the numerator 
$-1 + 2 \cos x+ \cos^2 x$, which will be $>0$ exactly on the countable union of intervals
$$( - \arccos(\sqrt{2}-1) + 2 k \pi, \arccos(\sqrt{2}-1) + 2 k \pi)$$
and this gives the solution for $x$.
It correlates with the answer of @Gonate: since $\tan ( \frac{1}{2}\cdot \arccos(\sqrt{2}-1)) = \sqrt{ \sqrt{2}-1}$.
$\arccos(\sqrt{2}-1)= 1.1437.. $ radians  or approximately $65.53^{\circ}$ degrees. 

The factor $e^{-1/(1+\cos x)}$ flattens away the singularities arising from the denominator around $(2k+1)\pi$. 
