Question

If tanA/2=±√1-e/1+e*tanB/2

So prove

CosB = cosA-e/1-ecosA

Answer 1

TanA/2=√(1-e)/(1+e) tanB/2

or, tanB/2=√(1+e)/(1-e) tanA/2

squaring both sides,
tan²B/2=(1+e)/(1-e) tan²A/2
or,
(1-tan²B/2)/(1+tan²B/2)={(1-e)-(1+e)tan²A/2}/{(1-e)+(1+e)tan²A/2}

[by dividendo-componendo method]
or,
cosB=(1-e-tan²A/2-etan²A/2)/(1-e+tan²A/2+etan²A/2)
or,
cosB={(1-tan²A/2)-e(1+tan²A/2)}/{(1+tan²A/2)-e(1-tan²A/2)}
or,
cosB=[{(1-tan²A/2)-e(1+tan²B/2)}/(1+tan²A/2)]/
[{(1+tan²A/2)-e(1-tan²A/2)}/(1+tan²A/2)]
or,
cosB=(cosA-e)/(1-ecosA) [∵, (1-tan²A/2)/(1+tan²A/2)=cosA] (Proved)