Answer
$\dfrac{- \sqrt 2}{2}$
Work Step by Step
The range of $\tan^{-1}{x}$ is $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$.
Here, $\tan^{-1}(-1)=\dfrac{-\pi}{4}$, because $\tan{(\dfrac{-\pi}{4})}= -1$ and $\dfrac{-\pi}{4}$ lies in the range of $\tan^{-1}{x}$ .
Therefore, we have:
$\sin [\tan^{-1}(-1)]=\sin ({\dfrac{-\pi}{4}})=\dfrac{- \sqrt 2}{2}$