<s>[INST] For $x > -1$ proof that $ \arctan x + \arctan\frac{1-x}{1+x} = \frac{\pi}{4} $ For $x > -1$ proof that $\arctan x + \arctan\dfrac{1-x}{1+x} = \dfrac{\pi}{4} $
I have no idea how to approach this, some kind of help would be greatly appreciated!
edit: Thank you all!
 [/INST] Let $\arctan(x) = a$. We then have
\begin{align}
\arctan(x) + \arctan \left(\dfrac{1-x}{1+x}\right) & = a + \arctan \left(\dfrac{1-\tan(a)}{1+\tan(a)}\right)\\
& = a + \arctan\left(\dfrac{\tan(\pi/4) - \tan(a)}{1+\tan(\pi/4) \cdot \tan(a)}\right)\\
& = a + \arctan \left(\tan(\pi/4-a)\right)
\end{align}
Note that since $x>-1$, we have $a = \arctan(x) \in \left(-\dfrac{\pi}4, \dfrac{\pi}2 \right)$. Hence, $$\dfrac{\pi}4 - a \in \left(-\dfrac{\pi}4, \dfrac{\pi}2 \right) \implies \arctan \left(\tan(\pi/4-a)\right) = \dfrac{\pi}4-a$$
This gives us
\begin{align}
\arctan(x) + \arctan \left(\dfrac{1-x}{1+x}\right)& = a + \arctan \left(\tan(\pi/4-a)\right)\\
& = a + \dfrac{\pi}4 - a = \dfrac{\pi}4
\end{align}
</s>