Write the following trigonometric identities problem as instructed:Maxwell's theory of electromagnetic waves established that electromagnetism...

We'll manage the right side of the identity and we'll transform
the difference from numerator and the sum from denominator, into
products.

sin2a - sin 2b =
2cos[(2a+2b)/2]*sin[(2a-2b)/2]

sin2a - sin 2b =
2cos[(a+b)]*sin[(a-b)] (1)

sin2a +sin2b =
2sin[(2a+2b)/2]*cos[(2a-2b)/2]

sin2a +sin2b = 2sin[(a+b)]*cos[(a-b)]
(2)

We'll divide (1) by (2) and we'll
get:

B/A =
2cos[(a+b)]*sin[(a-b)]/2sin[(a+b)]*cos[(a-b)]

B/A =
cos[(a+b)]*sin[(a-b)]/sin[(a+b)]*cos[(a-b)]

cos (a+b) = cosa*cos b -
sin a*sin b

sin (a-b) = sin a*cos b - sin b*cos
a

cos (a+b)*sin (a-b) = sin a*cos a*(cos b)^2 - sin b*cos b*(cos
a)^2 - sin b*cos b*(sin a)^2 + sin a*cos a*(sin b)^2

We'll group the
1st and the last terms and the middle terms and we'll get:

sin a*cos
a*[(cos b)^2 + (sin b)^2] - sin b*cos b*[(cos a)^2 + (sin a)^2]

But
(cos b)^2 + (sin b)^2 = 1 and (cos a)^2 + (sin a)^2 = 1

cos
(a+b)*sin (a-b) = sin a*cos a - sin b*cos b (3)

sin[(a+b)] = sin
a*cos b + sin b*cos a

cos[(a-b)] = cos a*cos b + sin
a*sin b

sin[(a+b)]*cos[(a-b)] = sin a*cos a*(cos b)^2 + sin b*cos
b*(sin a)^2 + sin b*cos b*(cos a)^2 + sin a*cos a*(sin
b)^2

sin[(a+b)]*cos[(a-b)] = sin a*cos a*[(cos b)^2 + (sin b)^2] +
sin b*cos b*[(cos a)^2 + (sin a)^2]

sin[(a+b)]*cos[(a-b)] = sin
a*cos a + sin b*cos b (4)

B/A =
(3)/(4)

B/A = (sin a*cos a - sin b*cos b)/(sin a*cos a + sin b*cos b
)

As we can notice, the option b) is the correct
option:

B/A = (sin a*cos a - sin b*cos
b)/(sin a*cos a + sin b*cos b )