if a cos 2 theta + bsin 2 theta =c has a and b as its solution,then prove that tan a +tan b = 2b/a+b

Acos 2theta+ bsin 2theta=c=> a (1-tan^2 theta /1+tan^2 theta)+ b(2tan theta/ 1+tan ^2 theta)=c=>(-a-c)tan^2theta + 2btantheta+ a-c=0=> tan a+ tan b= -2 b/(-a-c)=2b/a+cSince a and bare the roots of the equation and sum of roots is the negative ratio of coefficients of x by coefficients of x^2Manish kumar panda