Let $f \left(x\right) = {\cos }^{2}x e^{\tan x} , x \in \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$then

$f\left(x\right) = {\cos }^{2}x · e^{\tan x}$

$\Rightarrow f \text{'}\left(x\right) = e^{\tan x}.· (1 – \sin 2x) \ge 0$$\forall x\in \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$$\Rightarrow f\left(x\right)\uparrow$on $\left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$

$f "\left(x\right) = e^{\tan x}\left(–2 \cos 2x\right)+e^{\tan x} se{c}^{2}x\left(1-\sin 2x\right)$

$\therefore f "\left(x\right) = e^{\tan x }\left(\frac{2({\tan }^{2}x-1)}{1+{\tan }^{2}x}+(se{c}^{2}x-2\tan x)\right)=f\left(x\right) (\tan x – 1) ({\tan }^3 x – {\tan }^{2}x + 3 \tan x + 1)$

Now, let $g \left(x\right) = {\tan }^3 x – {\tan }^2 x + 3 \tan x + 1$

$g\text{'}\left(x\right) = (3 {\tan }^{2}x – 2 \tan x + 3) se{c}^{2}x > 0$$\forall x\in \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$

So, $g\left(x\right)$ is increasing function.

Also, $g\left(\frac{-\mathrm{\pi }}{4}\right) g\left(0\right) < 0$

So,$g\left(x\right) = 0$ has exactly one root in $\left(\frac{-\mathrm{\pi }}{2}\frac{\mathrm{\pi }}{2}\right)$,

At $x=\frac{\mathrm{\pi }}{4}$, $f \text{'}\left(x\right)$ has a local minima.