the problem is tan(x)+cot(x)=2csc(2x). I can get to the point where I have (2sin(x)cos(x))/(sin(x)cos(x)). I am not quite sure what to do beyond this point.
You should post this in a new thread.
sin(2x) = 2sin(x)cos(x)
cos(2x) = 1 - 2[sin(x)]^2 = 2[cos(x)]^2 - 1 = [cos(x)]^2 - [sin(x)]^2
so
sin(x)+cos(2x)-1 = sin(x) + 1 - 2[sin(x)]^2 - 1 = -2[sin(x)]^2 + sin(x) = sin(x) {-2sin(x) + 1}
cos(x)-sin(2x) = cos(x) - 2sin(x)cos(x) = cos(x) {1 - 2sin(x)}
Dividing the two cancels the common 1 - 2sin(x) leaving sin(x)/cos(x) = tan(x).
-Dan