1.
Put A in place of ?.
Multiply top and bottom by sinA-icosA:
(1+a)/(1-a)=(1+cosA+isinA)/(1-cosA-isinA)=
(1+cosA+isinA)(1-cosA+isinA)/((1-cosA)^2+sin^2(A)).
Numerator: 1-cos^2(A)+isinA(1+cosA+1-cosA)-sin^2(A)=
sin^2(A)-sin^2(A)+2isinA=2isinA.
Denominator: 1-2cosA+cos^2(A)+sin^2(A)=2-2cosA.
Numerator/denominator=isinA/(1-cosA).
But sinA=2sin(A/2)cos(A/2) and cosA=1-2sin^2(A/2),
so isinA/(1-cosA)=2isin(A/2)cos(A/2)/2sin^2(A/2)=icot(A/2).
Therefore, (1+a)/(1-a)=icot(A/2) QED
2.
v has been assumed to represent square root.
(x+iy)(x-iy)=x^2+y^2.
(x+iy)/(x-iy)=(x+iy)^2/(x^2+y^2)=
(ac+bd+ibc-iad)/((c^2+d^2)(x^2+y^2))
(x+iy)^2=x^2-y^2+2ixy=
(a+ib)/(c+id)=(a+ib)(c-id)/(c^2+d^2)=(ac+bd+ibc-iad)/(c^2+d^2).
So x^2-y^2=(ac+bd)/(c^2+d^2) and 2xy=(bc-ad)/(c^2+d^2).
Also, 1/(x+iy)=(x-iy)/(x^2+y^2)=
√((c+id)/(a+ib))=√((c+id)(a-ib)/(a^2+b^2))=
√((ac+bd+iad-ibc)/(a^2+b^2)).
More to follow...