tantan−11a+b+tan−1ba2+ab+1=
a
1a
b
1b
we have tantan−11a+b+tan−1ba2+ab+1 =tantan−11a+b+tan−1(a+b)−a1+(a+b)a =tantan−11a+b+tan−1(a+b)-tan−1a =tanπ2-tan−1a =tancot−1a =tantan−11a =1a