Let tan−1(31)=θ and tan−1(71)=ϕ
∴tanθ=31 and tanϕ=71
Now 2tan−1(31)+ tan−1(71)= 2θ+ϕ
and
tan(2θ+ϕ)
=1−tan2ϕtanϕtan2θ+tanϕ
=1−(1−tan2θ2tanθ)×tanϕ1−tan2θ2tanθ+tanϕ
=1−(1−9132)×711−9132+71
=(1−43×7143+71)×2828
=28−321+4
∴2θ+ϕ= 4π+kπ, for some k∈Z.
But tan−1(31)=θ and tan−1(71)=ϕ both lie between 0 and 4π.
∴2θ+ϕ=
∴2tan−1(31)+tan−1(71)=