1. The range of the function $f(x)=\sin x+\cos x+\tan x+$ $\arcsin x+\arccos x+\arctan x$ is $\qquad$ .
$=、 1 \cdot\left[-\sin 1+\cos 1-\tan 1+\frac{\pi}{4}, \sin 1+\cos 1+\tan 1+\frac{3 \pi}{4}\right]$.
Obviously, $f(x)=\sin x+\cos x+\tan x+\arctan x+\frac{\pi}{2}$, $x \in[-1,1]$.

Below we prove: $g(x)=\cos x+\tan x$ is an increasing function on $[-1,1]$.
Let $x_{1} 、 x_{2} \in[-1,1]$, and $x_{1}0$, thus,
$$
\frac{\cos \frac{x_{1}-x_{2}}{2}}{\cos x_{1} \cdot \cos x_{2}} \geqslant \frac{1}{\cos \frac{x_{1}-x_{2}}{2}} \geqslant 1 .
$$

Note that the equality cannot hold simultaneously, therefore,
$$
\frac{\cos \frac{x_{1}-x_{2}}{2}}{\cos x_{1} \cdot \cos x_{2}}>1 \text {. }
$$

Thus, $g\left(x_{1}\right)<g\left(x_{2}\right)$.
Hence $g(x)$ is an increasing function on $[-1,1]$.
Since $y=\sin x, y=\arctan x$ are increasing functions on $[-1,1]$, therefore, $f(x)$ is an increasing function on $[-1,1]$, and its range is $[f(-1), f(1)]$.