Cos (a + b) = 12 / 13, cos (a-b) = - 12 / 13, and A-B belongs to (PI / 2, PI), a + B belongs to (3pi / 2,2pi) find cos2a-cos2b

Cos (a + b) = 12 / 13, cos (a-b) = - 12 / 13, and A-B belongs to (PI / 2, PI), a + B belongs to (3pi / 2,2pi) find cos2a-cos2b

A-B belongs to (PI / 2, PI)
In the second quadrant, so sin (a-b) > 0
cos(a-b)=-12/13
So sin (a-b) = 5 / 13
In the same way
A + B in quadrant 4
cos(a+b)=12/13
Then sin (a + b) = - 5 / 13
therefore
cosacosb+sinasinb=-12/13
cosacosb-sinasinb=12/13
cosacosb=0,sinasinb=-12/13
sinacosb+cosasinb=-5/13
sinacosb-cosasinb=5/13
sinacosb=0,cosasinb=-5/13
From cosacosb = 0 to sinacosb = 0
Cosa and Sina cannot be 0 at the same time
So there must be CoSb = 0
sinasinb=-12/13
cosasinb=-5/13
CoSb = 0, so SINB is not equal to 0
So Sina / cosa = 12 / 5
sina=12/5*cosa
(sina)^2+(cosa)^2=1
So (COSA) ^ 2 = 25 / 169
cos2a-cos2b
=2(cosa)^2-1-2(cosb)^2+1
=2(cosa)^2-2(cosb)^2
=2*25/169-2*0
=50/169