Answer
$-\frac{7}{25}$
Work Step by Step
1. Let $tan^{-1}(-\frac{4}{3})=t$ (t in quadrant IV), we have $tan(t)=-\frac{4}{3}$ and $cos(t)=\frac{3}{5}$
2. $cos(2tan^{-1}(-\frac{4}{3}))=cos(2t)=2cos^2(t)-1=2(\frac{3}{5})^2-1=-\frac{7}{25}$
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