<s>[INST] Draw the function $y=\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^x-4}{e^x\sqrt{3}}}$ I'm not allowed to use differentiation for this. I know the domain and range of the arctan function. The function $\arctan:\mathbb{R}\rightarrow \left(-\frac{\pi}{2},\frac{\pi}{2}\right).$ Is my job to see what happens to the arguments $\frac{e^x-1}{\sqrt{3}}$ and $\frac{e^x-4}{e^x\sqrt{3}}$ in this range? I'm sorry for not having a better question but I'm stuck.
 [/INST] Note:If you find $f'$ when $f(x)=y=\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^x-4}{e^x\sqrt{3}}}$ , you will see
$f'(x)=0$ so $f(x)$ is a constant function 
$$f(x)=\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^x-4}{e^x\sqrt{3}}}=\\
\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{e^{-x}(e^x-4)}{\sqrt{3}}}=\\
\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{1-4e^{-x}}{\sqrt{3}}}\\$$I tis easier to work .
  $$f'(x)=\frac{\bigg(\dfrac{e^x-1}{\sqrt{3}}\bigg)'}{1+\bigg(\dfrac{e^x-1}{\sqrt{3}}\bigg)^2}-\frac{\bigg(\dfrac{1-4e^{-x}}{\sqrt{3}}\bigg)'}{1+\bigg(\dfrac{1-4e^{-x}}{\sqrt{3}}\bigg)^2}\\=
\frac{\frac{e^x}{\sqrt{3}}}{1+\frac{(e^x-1)^2}{3}}-\frac{\frac{4e^{-x}}{\sqrt{3}}}{1+\frac{(1-4e^{-x})^2}{3}}
$$ to simplify $\frac{\frac{4e^{-x}}{\sqrt{3}}}{1+\frac{(1-4e^{-x})^2}{3}} $  multiply by $\frac{e^{2x}}{e^{2x}}$ hence:
$$f'=\frac{\sqrt3 e^x}{4+e^{2x}-2e^x}-\frac{4\sqrt3 e^{-x}}{4+16e^{-2x}-8e^{-x}}\\=
\frac{\sqrt3 e^x}{4+e^{2x}-2e^x}-\frac{4\sqrt3 e^{-x}}{4(1+4e^{-2x}-2e^{-x}}.\frac{e^{2x}}{e^{2x}}\\=
\frac{\sqrt3 e^x}{4+e^{2x}-2e^x}-\frac{\not4\sqrt3 e^{+x}}{\not4(1.e^{2x}+4e^{-2x+2x}-2e^{-x+2x}}.\frac{e^{2x}}{e^{2x}}\\=
\dfrac{\sqrt3 e^x}{4+e^{2x}-2e^x}-\frac{\sqrt3 e^x}{4+e^{2x}-2e^x} \\=0 \checkmark$$ so $\forall x \in \mathbb{R} :f'=0\implies f(x)=const$ now check a number in $f(x)$
forexample $x=0$
$$f(0)=\arctan{\frac{1-1}{\sqrt{3}}}-\arctan{\frac{1-4}{1.\sqrt{3}}}=0-(-\frac{\pi}{3})=\frac{\pi}{3}$$
A trig way $$\arctan{\frac{e^x-1}{\sqrt{3}}}-\arctan{\frac{1-4e^{-x}}{\sqrt{3}}}=a-b \\ \tan(a-b)=\frac{\tan a -\tan b}{1+\tan a . \tan b}=\\\frac{\frac{e^x-1}{\sqrt{3}} -\frac{1-4e^{-x}}{\sqrt{3}}}{1+\frac{e^x-1}{\sqrt{3}} . \frac{1-4e^{-x}}{\sqrt{3}}}=\sqrt{3}\underbrace{\frac{e^x-1-1+4e^{-x}}{3+(e^x-1).(1-4e^{-x})}}_{1}=\\\sqrt 3 \implies \tan(a-b)=\arctan \sqrt3=\frac{\pi}{3}|$$
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