1.

`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)]`

Answer» `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))]`
`= 2[(1 + tan^2 ((theta - phi)/2) + 1 - tan^2((theta - phi)/2))/(1 + tan^2((theta - phi)/2))]`
`a^2 + b^2 = 4/(1 + tan^2((theta - phi)/2))`
`1 + tan^2 ( (theta - phi)/2) = 4/(a^2 + b^2)`
`tan( (theta - phi)/2) = +- sqrt(4/(a^2 + b^2) - 1)`
`= +- sqrt((4 - a^2 - b^2)/(a^2 + b^2))`
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