[tan(x-Ï€/4)]=[1 tan(x)]/[1-tan(x)]

I think the identity you are asking about is this:


\tan (x+\frac {\pi} 4 ) = \frac {1+\tan (x)}{1-\tan (x)}


Start by converting the tangent functions to sin/cos. Then apply the identities sin(a+b) = sin(a)cos(b)+cos(a)sin(b) and cos(a+b) = cos(a)cos(b)-sin(a)sin(b).