Establishing Trig identities

I've tried everything I can think of the closest I ever got to the answer is on one of the tries when I went to the x,y,r system and got 1/y.

Here's what I was able to get: 

1. cos@tan@ + cos@cot@ = csc@
2. cos@(sin@/cos@) + cos@(cos@/sin@) = csc@
3. sin@ + cos^2@/sin@ = csc@
4. sin@(sin@ + cos^2@/sin@) = sin@csc@
5. sin^2@ + cos^2@ = sin@(1/sin@)
6. 1 = 1

In step 1 I distributed.  In step 2 I rewrote both tan and cot in terms of sin and cos.  In step 3 I simplified.  In step 4 I multiplied both sides by sin in order to get rid of the denominator under cos^2.  In step 5 I simplified, and rewrote csc in terms of sin on the right side of the eq.  Finally, by the Pythagorean identity, the left side equals 1, while the right side equals 1 as well.

Hope this helps.