Alternative solution:
To prove:
arccos{(cos(a)+cos(b))/(1+cos(a)cos(b))}=
2arctan{tan(a/2)tan(b/2)}.
Let cos(A)=(cos(a)+cos(b))/(1+cos(a)cos(b)).
Let tan(B)=tan(a/2)tan(b/2).
So essentially we need to prove A=2B.
Let t=tan(θ/2), so we can expand the trig identity:
tan(θ)=2t/(1-t²), so t²tan(θ)+2t-tan(θ)=0.
So, t²+2tcot(θ)=1.
Completing the square:
t²+2tcot(θ)+cot²(θ)=1+cot²(θ)=csc²(θ).
(t+cot(θ))²=csc²(θ), t=±csc(θ)-cot(θ).
So tan(θ/2)=(±1-cos(θ))/sin(θ).
If we keep θ positive, tan(θ/2)=(1-cos(θ))/sin(θ).
So tan(a/2)=(1-cos(a))/sin(a) and tan(b/2)=(1-cos(b))/sin(b).
tan(a/2)tan(b/2)=((1-cos(a))/sin(a))((1-cos(b))/sin(b)).
sin(a)sin(b)=½(cos(a-b)-cos(a+b)),
cos(a)cos(b)=½(cos(a-b)+cos(a+b)).
If A=2B, then secA=sec(2B).
secA=(1+cos(a)cos(b))/(cos(a)+cos(b)).
tan(2B)=2tanB/(1-tan²B)=2tan(a/2)tan(b/2)/(1-tan²(a/2)tan²(b/2)).
2tanB=2tan(a/2)tan(b/2)=2(1-cos(a))(1-cos(b))/(sin(a)sin(b)).
tan²B=tan²(a/2)tan²(b/2)=(1-cos(a))²(1-cos(b))²/((1-cos²(a))(1-cos²(b)).
tan²B=((1-cos(a))(1-cos(b)))/((1+cos(a))(1+cos(b))).
1-tan²B=((1+cos(a))(1+cos(b))-(1-cos(a))(1-cos(b)))/((1+cos(a))(1+cos(b)))=
2(cos(a)+cos(b))/((1+cos(a))(1+cos(b))).
tan(2B)=sin(a)sin(b)/(cos(a)+cos(b)).
sec²(2B)=1+tan²(2B)=((cos(a)+cos(b))²+sin²(a)sin²(b))/(cos(a)+cos(b))².
Replace sin²(a) with 1-cos²(a) and sin²(b) with 1-cos²(b):
(cos(a)+cos(b))²+(1-cos²(a))(1-cos²(b))=
cos²(a)+2cos(a)cos(b)+cos²(b)+1-cos²(a)-cos²(b)+cos²(a)cos²(b)=(1+cos(a)cos(b))².
sec²(2B)=(1+cos(a)cos(b))²/(cos(a)+cos(b))².
sec(2B)=(1+cos(a)cos(b))/(cos(a)+cos(b))=secA. QED