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!
 A: An easy way to show this without trigonometry is to calculate the derivative of the LHS and show that it is zero. Then plug in some $x$, say $x=0$, to arrive at the desired result.
A: 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}
A: Hint: Use the probably familiar formula 
$$\tan(u+v)=\frac{\tan u+\tan v}{1-\tan u\tan v}.$$
To use the formula for your problem, take the tangent of both sides. Then you will need to do some algebra.
Remark: If you need to prove that the formula mentioned above holds, write down the formulas for $\sin(u+v)$ and $\cos(u+v)$. Divide, and in the expression you get, divide top and bottom by $\cos u\cos v$.
A: $$
\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}
$$
for $(x,y)\in\text{a suitable set}$, because of the formula for the tangent of a sum.
