>>896
a(n)=tanθ とおくと
a(n+1)=(1+tanθ)/(1-tanθ)=(cosθ+sinθ)/(cosθ-sinθ)
=√2sin(θ+π/4)/{√2cos(θ+π/4)}=tan(θ+π/4)

a(1)=√3=tan(π/3) だから
a(n)=tan{(π/3)+(n-1)*(π/4)}=tan{(nπ/4)+(π/12)}