## Calculus (3rd Edition)

$-\sqrt 3$
Since $\sec^{-1} x$ is an even function, then we have $\sec^{-1}(-2)=\sec^{-1} 2$. Now, assume that $\sec\theta =\frac{2}{1}$, then by solving the triangle, we have: $$\tan \theta=\sqrt 3.$$ Now, since $\theta =\sec^{-1}2$, we have $$\tan(\sec^{-1}(-2) )=\tan\theta=-\sqrt 3 .$$ We know that tangent is negative because secant is negative in the 2nd and 3rd quadrants. We exclude the 3rd quadrant because it is outside the range of arcsecant. We know that tangent is negative in the 2nd quadrant, thus the answer must be negative.