Answer
The identity is verified.
$tan(\frac{\pi}{4}-θ)=\frac{1-tan~θ}{1+tan~θ}$
Work Step by Step
$tan(u-v)=\frac{tan~u-tan~v}{1+tan~u~tan~v}$
$tan(\frac{\pi}{4}-θ)=\frac{tan~\frac{\pi}{4}-tan~θ}{1+tan~\frac{\pi}{4}~tan~θ}=\frac{1-tan~θ}{1+(1)tan~θ}=\frac{1-tan~θ}{1+tan~θ}$
