`sin theta + sin phi = a`,`cos theta + cos phi=b` then prove `tan [(theta -phi)/2]=+-sqrt[(4-a^2-b^2)/(a^2+b^2)]`

Explanation

`sqrt((4-(a^2 + b^2))/(a^2 + b^2))`

`= sqrt(4/(a^2 + b^2) -1)`

let `a^2 + b^2 = (sin theta + sin phi)^2 + (cos theta + cos phi)^2`

`= sin^2 theta + sin^2 phi + 2 sin theta sin phi + cos^2 theta + cos^2 phi + 2 cos theta cos phi`

`1 + 1 + 2[cos theta cos phi + sin theta sin phi]`

`= 2 + 2 cos ( theta - phi)`

`= 2[ 1+ cos( theta - phi)]`

`= 2[ 1 + (1- tan^2((theta - phi)/2))/(1 + tan^2((theta - phi)/2))]`

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