Question 818520

Given
sin^2x-sin^2y=cos^2y-cos^2x

Solution: 
Pythagorean Identity
   cos^2x + sin^2y = 1
   sin^2Ø = 1 - cos^2Ø
 
 sin^2x-sin^2y=cos^2y-cos^2x
 (1-cos^2x)-(1 - cos^2y)=cos^2y-cos^2x
  1 -cos^2x -1 + cos^2y =cos^2y-cos^2x
  1 -1 -cos^2x + cos^2y =cos^2y-cos^2x
       -cos^2x + cos^2y =cos^2y-cos^2x

        cos^2y - cos^2x =cos^2y-cos^2x