Prove that sina-sin3a+sin5a-sin7a/cosa-cos3a-c os5a+cos7a= cot2a

Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
    = -2cos4A*sin3A + 2cos4A*sinA
    = 2cos4A(sinA - sin3A)
    = 2cos4A*2cos(2A)sin(-A)
    = -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)
   = 2cos4A*cos3A - 2cos4A*cosA
   = 2cos4A(cos3A - cosA)
   = 2cos4A*(-2)sin2A*sinA
   = -4cos4A*sin2A*sinA

Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
       = cos2A/sin2A
      = cot2A

This verifies the identity.

Aand b are correctly balanced reactions. they have the same number of atoms in both sides of the reaction