Question

Plzzz help me.....
If tan A/2=√1-e/1+e tan B/2...... Then prove that cosA= cosA-e/1-e cosA

Answer 1

tanθ/2=√(1-e)/(1+e) tanФ/2

or, tanФ/2=√(1+e)/(1-e) tanθ/2

squaring both sides, 

tan²Ф/2=(1+e)/(1-e) tan²θ/2

or, (1-tan²Ф/2)/(1+tan²Ф/2)={(1-e)-(1+e)tan²θ/2}/{(1-e)+(1+e)tan²θ/2}

[by dividendo-componendo method]

or, cosФ=(1-e-tan²θ/2-etan²θ/2)/(1-e+tan²θ/2+etan²θ/2)

or, cosФ={(1-tan²θ/2)-e(1+tan²θ/2)}/{(1+tan²θ/2)-e(1-tan²θ/2)}

or, cosФ=[{(1-tan²θ/2)-e(1+tan²θ/2)}/(1+tan²θ/2)]/

                                                   [{(1+tan²θ/2)-e(1-tan²θ/2)}/(1+tan²θ/2)]

or, cosФ=(cosθ-e)/(1-ecosθ) [∵, (1-tan²θ/2)/(1+tan²θ/2)=cosθ] (Proved)

hope it helps...