Using a sum-to-product identity, we have
sin x + sin y = 2sin[(x + y)/2] cos[(x - y)/2]
And
cos x + cos y = 2cos[(x + y)/2]cos[(x - y)/2]
So
[sin x + sin y] / [cos x + cos y] =
[2sin[(x + y)/2] cos[(x - y)/2] ] / [2cos[(x + y)/2]cos[(x - y)/2] ] =
sin [(x + y)/2] / cos[ (x + y)/2 ] and since (x + y)/2 is just an angle, θ, then
sinθ / cosθ = tanθ = tan[(x + y}/2]
The other part can be proved similarly by noting that
sinx - sin y = 2cos[(x + y)/2] sin[(x - y)/ 2]
