Question

[sin(A+3B)+sin(3A+B)]/[sin2A+sin2B]=2cos(A+B)

Answer 1

{sin(A +3B) + sin(3A +B)}/{sin2A +sin2B}


use formula,
sinA + sinB = 2sin(A+B)/2.cos(A-B)/2
sin2A = 2sinA.cosA
cos(-@) = cos@

now,

{2sin(A + 3B + 3A + B)/2.cos(A +3B-3A-B)/2 }/{2sin(A + B).cos(A -B)}

=> sin(2A +2B).cos(B - A)/sin(A + B).cos(A -B)

=>2sin(A +B).cos(A + B).cos(B -A)/sin(A + B).cos(A - B)

=> 2cos( A + B)

Answer 2

Answer:

Step-by-step explanation: