Sketching $y = \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{e^x-4}{\sqrt3e^x}\right)$ The exercise is to graph this function below.
$$y = \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{e^x-4}{\sqrt3e^x}\right)$$
If the exercise were just one trigonometric function, then I would just put the expression inside arctangent in argtangent's domain and moreover put $y$ in the range of which the argtangent is defined. However there are two trigonometric functions in the equation above, and I do not know where I would start to able to sketch it. 
I also want to add that I solved the equation algebraically by taking the tangent of both sides and using the subtaction of two angle for tangent
$$ \tan(x-y)=\frac{\tan\ x + \tan\ y}{1 + \tan\ x*\tan\ y}$$
The equation simplified to $\tan\ y = \sqrt3$, which gives us $y = \frac{\pi}{3}$. Does this $y$ value help me sketch the function in anyway?  
 A: Note that the arctan
has an implied
$+k\pi$
since
$\tan(x+k\pi)
=\tan(x)$,
so you have to be careful.
For example,
if you naively evaluate
$\arctan(1)+\arctan(2)+\arctan(3)$
you get
$\arctan(1)+\arctan(2)
=\arctan(\frac{1+2}{1-1\cdot 2})
=\arctan(-3)
$
so
$\arctan(1)+\arctan(2)+\arctan(3)
=\arctan(-3)+\arctan(3)
=0$,
but the actual sum,
using principal values,
is $\pi/2$.
For this case,
when $x > 0$,
$\begin{array}\\
y 
&= \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{e^x-4}{\sqrt3e^x}\right)\\
&= \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{1}{\sqrt3}-\frac{4}{\sqrt3e^{x}}\right)\\
&= t_1-t_2\\
\end{array}
$
Since $e^x > 1$,
$\frac{e^x-1}{\sqrt3} > 0$
so
$0 < t_1 < \pi/2$
and
$\frac{1}{\sqrt3}-\frac{4}{\sqrt3e^{x}}
\gt \frac{1-4/3}{\sqrt3}
=-\frac1{3\sqrt{3}}
$
so
$-\arctan(\frac1{3\sqrt{3}})
\le t_2
\lt \arctan(1/\sqrt{3})
=\pi/6
$.
Therefore
$-\pi/6
\lt t_1-t_2
\lt \pi/2-\arctan(\frac1{3\sqrt{3}})
\approx \pi/2-0.190
$
and,
since
$\lim_{x \to \infty} t_1 = \pi/2$
and
$\lim_{x \to \infty} t_2 = \pi/6$,
we have
$\lim_{x \to \infty} t_1-t_2 = \pi/3$.
For $x < 0$
you can do a
similar analysis
using
$0 < e^x < 1$.
