An exam question concerning trig identities?

Question

Use the identities for (sin A + sin B) and (cos A + cos B) to prove that:

(sin2x + sin2y)/(cos2x + cos2y) = tan(x + y)

A notice to the pretentious, smug idiot, who assumed that this is a real exam where I somehow have access to a computer, despite it being nearly midnight (I don't understand how it's actually possible to believe something as patently false as this), and choosing instead to hurl petty ego boosting insults:

This is a practice exam that you do for homework (if you hadn't already worked that out by now). Pretty much every question on here is homework help. I have tried many times to answer this question but simply cannot do it. So I ask you for help.

Answer 1

The identities are:

(sinA + sinB) = 2sin((A + B)/2)cos((A - B)/2)
(cosA + cosB) = 2cos((A + B)/2)cos((A - B)/2)

Therefore,

(sin2x + sin2y) = 2sin(x + y)cos(x - y)
(cos2x + cos2y) = 2cos(x + y)cos(x - y)

(sin2x + sin2y) / (cos2x + cos2y) = 2sin(x + y)cos(x - y) / 2cos(x + y)cos(x - y) = sin(x + y) / cos(x + y) = tan(x + y), QED

Answer 2

The fact that you're relying on strangers to answer an exam question, be they right or convincing or not on both counts, is appalling.
The fact that you're too lazy to figure it out is sad.
The fact that you're so willing to cheat is disgusting.