Answer
$\frac{1519}{1681}$
Work Step by Step
Step 1. Let $u=tan^{-1}\frac{9}{40}$, we have $tan(u)=\frac{9}{40}$ and $sin(u)=\frac{9}{\sqrt {40^2+9^2}}=\frac{9}{41}$
Step 2. $cos(2tan^{-1}\frac{9}{40})=cos(2u)=1-2sin^2(u)=1-2\times(\frac{9}{41})^2=\frac{1519}{1681}$
