Prove that sin (45°+ A)cos (45°- B) + cos (45°+ A)sin (45°- B) = cos (A-B)

Solution:

To prove this, we have to use Sum & Difference Identities.

LHS = sin (45°+ A)cos (45°- B) + cos (45°+ A)sin (45°- B)

= sin [(45°+ A)+(45°- B)]…… {∵ sin (A + B) = sinA cosB + cosA sinB, A= 45°+ A, B= 45°- B }

= sin [90°+(A-B)]…. {∵ sin (90+θ) = cos θ}

= cos (A-B)

= RHS