If $f: \mathbb{R} \to \mathbb{R}$ is continous, $g(x) = 5^\frac{x+1}{x-1} + \cot(x)$, prove that there is $a$ $\in$ $Dg$, such that: $$f(a) = g(a)$$
Is it enough to say that, because $\lim_{\frac{\pi}{2}-0} g(x) = +\infty$ and $\lim_{-\frac{\pi}{2}+0} g(x) = -\infty$, and $f$ is continous on $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so $f$ bounded on that segment, there must be $c, d$ in that segment such that $g(c)-f(c)$ is positive and $g(d)-f(d)$ is negative, so by IVT there is $a$ between $c$ and $d$ such that $f(a)-g(a)=0$ .. ?
If that's true, how to put it formally, if it's not, why ?
Your argument is correct. Let $$m:=\min_{x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]}f(x),\qquad M=\max_{x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]}f(x)$$ which exist finite because $f$ is continuous. Now since $\lim_{x\to -\pi/2^+}g(x)=-\infty$ and $\lim_{x\to \pi/2^-}g(x)=+\infty$, there is $x_1,x_2\in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ with $g(x_1)<m$ and $g(x_2)>M$. Thus by construction of $m$ and $M$ we have $g(x_2)-f(x_2)>0$ and $g(x_1)-f(x_1)<0$.
By the intermediate value theorem, you are done.