<s>[INST] Why $\frac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \frac{\pi}{4}$? Since
$$\dfrac{d}{dx} \left( \dfrac{1}{2} \arctan(x) \right) = \dfrac{d}{dx} \left( \arctan(x-\sqrt{x^2+1}) \right) $$
then the format of their graphs are the same, but separated by a constant, which is $\dfrac{\pi}{4}$. That is,
$$\dfrac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \dfrac{\pi}{4} $$
My question is: why is this constant and what procedure is done to discover it.
 [/INST] I'm going to use $y$ instead of $x$. 
Let $P(\theta) = (1, y)$. Then  $$\theta = \arctan(y).$$ 

Then 
$$\sin \theta = \dfrac{y}{\sqrt{1+y^2}} 
  \qquad \text{and} \qquad
  \cos \theta = \dfrac{1}{\sqrt{1+y^2}}$$
Then \begin{align}
   \tan \bigg( \frac 12 \theta - \frac{\pi}{4} \bigg)
   &= \dfrac{\tan \frac 12 \theta - \tan \frac{\pi}{4}}
            {1 + \tan(\frac 12 \theta) \tan(\frac{\pi}{4})}\\
   &=\dfrac{\left(\dfrac{\sin \theta}{1+\cos \theta}-1 \right)}
           {\left( 1+\dfrac{\sin \theta}{1+\cos \theta} \right)} \\
   &= \frac{\sin \theta - 1 - \cos \theta}
           {1 + \cos \theta + \sin \theta} \\
   &= \frac{\left( \dfrac{y}{\sqrt{1+y^2}} - 1 - \dfrac{1}{\sqrt{1+y^2}} \right)}
           {\left( 1 + \dfrac{1}{\sqrt{1+y^2}} + \dfrac{y}{\sqrt{1+y^2}} \right)} \\
   &= \frac{y - \sqrt{1+y^2} - 1}{\sqrt{1+y^2} + 1 + y} \\
   &= \frac{(y - 1) - \sqrt{1+y^2}}{(y + 1) + \sqrt{1+y^2}} \cdot
      \frac{(y + 1) - \sqrt{1+y^2}}{(y + 1) - \sqrt{1+y^2}} \\
   &= \dfrac{2y^2 - 2y\sqrt{1+y^2}}{2y} \\
   &= y - \sqrt{1+y^2}
\end{align}
Since $\tan \bigg( \dfrac 12 \theta - \dfrac{\pi}{4} \bigg) = y - \sqrt{1+y^2}$,
then $$\dfrac 12 \theta - \dfrac{\pi}{4} = \arctan\left(y - \sqrt{1+y^2}\right).$$
In other words
$$\dfrac 12 \arctan(y) = \arctan\left(y - \sqrt{1+y^2}\right) + \dfrac{\pi}{4}.$$
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