`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.
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