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.