Question

(sina+sinb)/(sina-sinb)+(cosb-cosa)/(cosb+cosa)=2 /2 sin^2a-1

Answer 1

cos A - cos B = -2sin((A+B)/2)sin((A-B)/2) 
sin A - sin B = 2cos((A+B)/2)sin((A-B)/2) 
sin2 Ө + cos2 Ө = 1 

Then: 
(cos A - cos B)2 = [-2sin((A+B)/2)sin((A-B)/2)]2 = 4sin2(A+B)/2 * sin2(A-B)/2 
(sin A - sin B)2 = [2cos((A+B)/2)sin((A-B)/2)]2 = 4cos2(A+B)/2 * sin2(A-B)/2 

(cosA - cosB)2 + (sinA - sinB)2 = 4sin2(A+B)/2 * sin2(A-B)/2 + 4cos2(A+B)/2 * sin2 (A-B)/2 

Then, factor out 4sin2(A-B)/2 since it is common for both addends. 
It becomes: 4sin2(A-B)/2 * [sin2(A-B)/2 + cos2(A-B)/2] 

Recall that sin2Ө + cos2Ө = 1. This is similar to sin2(A-B)/2 + cos2(A-B)/2], where the Ө here is (A-B)/2. Therefore, sin2(A-B)/2 + cos2(A-B)/2 = 1. 

So, 4sin2(A-B)/2 * [sin2(A-B)/2 + cos2(A-B)/2] = 4sin2(A-B)/2 * 1 = 4sin2(A-B)/2