If `tan(alpha/2)=sqrt((a-b)/(a+b)) tan(beta/2)` prove that `cosalpha=(acosbeta+b)/(a+bcosbeta)`

Explanation

`L.H.S. = cos theta = (1-tan^2(theta/2))/(1+tan^2(theta/2))`

`=(1-((a-b)/(a+b)tan^2(phi/2)))/(1+((a-b)/(a+b)tan^2(phi/2)))`

`=(1-((a-b)/(a+b)sin^2(phi/2)/cos^2(phi/2)))/(1+((a-b)/(a+b)sin^2(phi/2)/cos^2(phi/2)))`

`=((a+b)cos^2(phi/2)-(a-b)sin^2(phi/2))/((a+b)cos^2(phi/2)+(a-b)sin^2(phi/2))`

`=(a(cos^2(phi/2) - sin^2(phi/2))+b(cos^2(phi/2) + sin^2(phi/2)))/(a(cos^2(phi/2) + sin^2(phi/2))+b(cos^2(phi/2) - sin^2(phi/2)))`

`=(acosphi+b)/(a+bcosphi)`

`=R.H.S.`