Prove that $e^x \tan x$ is strictly monotonically increasing on $(-\pi/2, \pi/2)$ I want to prove that $f: (-\pi/2,\pi/2) \to \mathbb{R}: x \mapsto e^x \tan x$ is strictly monotonically increasing on the given interval $D = (-\pi/2,\pi/2)$. I'd do this with the condition that
$$
f'(x) &gt; 0 \, \forall x \in D \Rightarrow f \text{ strictly mono. incr.} 
$$
See Q&amp;A answer below. It's a trivial question, really, but I already wrote it all out, so here it goes.
So, after writing my question down, I already found the answer myself, so this is gonna be a Q&amp;A answer.

Taking the derivative, we have
$$
\frac{d}{dx}f(x) = e^x \left( \tan x + \frac{1}{\cos^2 x}\right)
$$
Now, this product is obviously bigger than 0 if both factors are positive or both are negative. $e^x$ is always positive for $x \in \mathbb{R}$, so that means we must show that $(\tan x + \frac{1}{\cos^2 x}$) is positive.

This is where I was stuck. I then did the following steps:
$$
\begin{align}
\tan x + \frac{1}{\cos^2 x} &amp;&gt;0 \\
\tan x &amp;&gt; -\frac{1}{\cos^2 x} \\
\frac{\sin x}{\cos x} &amp;&gt; -\frac{1}{\cos^2 x} \\
\sin x \cdot \cos x &amp;&gt; -1
\end{align}
$$
Now we know that for $x \in D, 0&lt;\cos x \leq 1$ and $-1 &lt; \sin x &lt; 1$, and the product of these will always be $&gt;-1$. Q.E.D.
by Maclaurin Series we have:
  e^x=1+x/1!+x^2/2!+x^3/3!+⋯,-∞
  
  from this if 0 &lt; x &lt; y then e^x &lt; e^y and it follows that e^(-y) &lt; e^(-x). hence e^x is strictly increasing on the whole real axis.