If sin x + sin y = a and cos x + cos y = b, then tan^2((x+y)/2) + tan^2((x-y)/2) =?
Correct: D
sin x + sin y = a = 2 sin((x+y)/2)cos((x-y)/2). cos x + cos y = b = 2 cos((x+y)/2)cos((x-y)/2). Therefore, a/b = tan((x+y)/2). Then a^2+b^2 = 4cos^2((x-y)/2).
tan^2((x-y)/2) + tan^2((x+y)/2) = tan^2((x+y)/2) + (1-cos(x-y))/(1+cos(x-y)). Since a^2+b^2 = 4cos^2((x-y)/2) then cos^2((x-y)/2) = (a^2+b^2)/4. cos(x-y) = 2cos^2((x-y)/2) -1 = 2((a^2+b^2)/4)-1 = (a^2+b^2)/2-1 =(a^2+b^2-2)/2. Hence tan^2((x-y)/2) = (1-((a^2+b^2-2)/2))/(1+((a^2+b^2-2)/2)) = (4-a^2-b^2)/(a^2+b^2+2). Tan((x+y)/2) = a/b so tan^2((x+y)/2) = a^2/b^2, so (a^2/b^2) + (4-a^2-b^2)/(4+a^2+b^2).
Also 4-(a^2+b^2) = 4(1-cos^2((x-y)/2)) = 4sin^2((x-y)/2). Also 4+(a^2+b^2) = 4(1+cos^2((x-y)/2)). The answer can be found from solving.