If `sinalpha+sinbeta` and `cosalpha+cosbeta=b ,` prove that `tan(alpha-beta)/2=+-sqrt((4-a^2-b^2)/(a^2+b^2))` .

Explanation

Given `sin alpha+sin beta=alpha`
and `cos alpha+cos beta=b`
Now `(cos alpha+cos beta)^(2)+(sin alpha+sinbeta)^(2)=b^(2)+alpha^(2)`
or `cos^(2)alpha+cos^(2)beta+2cos alpha cos beta+sin^(2)alpha+sin^(2)beta`
`+2sinalpha sin beta=b^(2)+alpha^(2)`
or `(cos^(2)alpha+sin^(2)alpha)+(cos^(2)beta+sin^(2)beta)+2(cos alpha cos beta+sin alpha sin beta)=alpha^(2)+b^(2)`
or `2+2cos(alpha-beta)=alpha^(2)+b^(2)`
or `-cos(alpha-beta)=(alpha^(2)+beta^(2)-2)/(2)`
Now, `tan""(alpha-beta)/(2)+-sqrt((1-cos (alpha-beta))/(1+cos(alpha-beta)))`
`=+-sqrt(((1-(a^(2)+b^(2)-2))/(2))/((1+(a^(2)+b^(2)-2))/(2)))=+-sqrt((4-a^(2)-b^(2))/(a^(2)+b^(2)))`