`tan(pi/4-theta)=(1-tan(theta))/(1+tan(theta))`

we will use the following formula to prove the identity,

`tan(A-B)=(tanA-tanB)/(1+tanAtanB)`

LHS=`tan(pi/4-theta)`

`=(tan(pi/4)-tan(theta))/(1+tan(pi/4)tan(theta))`

plug in the value of tan(pi/4)=1,

`=(1-tan(theta))/(1+tan(theta))`

LHS=RHS, Hence proved.