Where do I start on this Trigonometry verification?

Verify    (cos(x + y)) / (cos(x) + sin(y)) = (cos(x) - sin(y)) / (cos(y - x))

step 1

\(\frac{cos(x + y)}{cos(x) + sin(y)} \stackrel{?}{=} \frac{cos(x) - sin(y)}{cos(y - x)}\)

step 2

\([cos(x + y)][cos(y - x) ]\stackrel{?}{=} [cosx - siny][cosx+ siny]\\\)

Step 3

cos is an even funtion so cos(y-x)=cos (-(y-x))

\([cos(x + y)][cos(x - y) ]\stackrel{?}{=} [cosx - siny][cosx+ siny]\\ \)

step4

\(let A=x+y\;\;\;B=x-y\\ LHS=cosAcosB\\ LHS=[(cosAcosB+sinAsinB)+(cosAcosB-sinAsinB)]*0.5\\ LHS=[(cos(A+B))+(cos(A-B))]*0.5\\ LHS=[(cos(x+y+x-y))+(cos(x+y-(x-y)))]*0.5\\ LHS=[(cos(2x))+(cos(2y))]*0.5\\ LHS=\frac{cos(2x)+cos(2y)}{2}\)

step 5

so we have

\(\frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [cosx - siny][cosx+ siny]\\ \frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [cos^2x - sin^2y]\\ \)

step 6

As an aside:

\(cos(2y)=cos^2y-sin^2y\\cos(2y)=1-2sin^2y\\sin^2y=\frac{1-cos(2y)}{2}\\ and\;\; likewise\;\;\\ cos^2x=\frac{1+cos2x}{2} \)

step 7

\(\frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [cosx - siny][cosx+ siny]\\ \frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [cos^2x -sin^2y]\\ likewise\\ \frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [\frac{cos(2x)+1}{2} - \frac{1-cos(2y)}{2}]\\ \frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} \frac{cos(2x)+cos(2y)}{2} \\ \)

Step 8

Left hand side and right hand side are identical so the equality is verified.