prove that:

cos2Acos2B + sin²(A - B) - sin²(A + B) = cos2(A + B)

sin²(A-B) - sin²(A+B) =sin(A+A)sin(-B-B) [using formula ,sin²a-sin²b=sin(a-b)sin(a+b)]

=sin(-2B)sin2A=-sin2Asin2B

Now LHS becomes

= cos2Acos2B-sin2Asin2B  [using formula ,cosacosb - sinasinb =cos(a+b)]

=cos(2A+2B)=cos2(A+B) = RHS 

hence proved

Cos2ACos2B+Sin^2(A-B)-Sin^2(A+B)=Cos2(A+B) Solution:- L.H.S:-Cos2ACos2B+Sin^2(A-B)-Sin^2(A+B) =Cos2ACos2B+Sin(A-B+A+B)*Sin(A-B-A-B) ["a^2-b^2=(a+b)(a-b)"] =Cos2ACos2B+Sin2A*Sin(-2B) =Cos2ACos2B-Sin2ASin2B =Cos2(A+B) ["Cos(A+B)=CosACosB-SinASinB"] Thank You Wish You A Happy Learning Trignometry.