1. $\arctan x+\arctan \frac{1-x}{1+x}=(x>-1)$.

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1. $\arctan x+\arctan \frac{1-x}{1+x}=(x>-1)$.
1. $\frac{\pi}{4}$.
From $x>1$ we know $-\frac{\pi}{2}<\arctan x+\arctan \frac{1-x}{1+x}<\pi, \tan \left(\arctan x+\arctan \frac{1-x}{1+x}\right)=$ $\frac{x+\frac{1-x}{1+x}}{1-x\left(\frac{1-x}{1+x}\right)}=1$, so $\arctan x+\arctan \frac{1-x}{1+x}=\frac{\pi}{4}$.