Answer
$\frac{4}{3}$
Work Step by Step
1. Let $sin^{-1}(\frac{4}{5})=u$ (quadrant I), we have $sin(u)=\frac{4}{5}, tan(u)=\frac{4}{3}$
2. Let $cos^{-1}(1)=v$ (x-axis), we have $v=0$
3. Thus $tan(u+v)=tan(u)=\frac{4}{3}$
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