If 0<θ,ϕ<π/2 and x=∑n=0∞ sin2nϕ, y=∑n=0∞ cos2nθ and z=∑n=0∞ cosn(θ+ϕ)cosn(θ−ϕ), then
xyz+1=yz-zx
xyz-1=yz+zx
xyz-xy=yz-zx
xyz+1=yz+zx
We have x=11−sin2ϕ=1cos2ϕ,y=11−cos2θ=1sin2θ,
Also, cos(θ+ϕ)cos(θ−ϕ)=cos2ϕ−sin2θ=1x−1yAs, 0<cos2ϕ<1,0<sin2θ<1,−1<cos2ϕ−sin2θ<1⇒ −1<1x−1y<1Thus, z=∑n=0∞ 1x−1yn=11−1x−1y=xyxy−y+x⇒ z(xy−y+x)=xy or xyz−xy=yz−xz