prove that cotx/(cosecx+1)+(cosecx+1)/cotx=2 secx

`csc^2(x) =
1+cot^2(x)`

`(csc(x)+1)(csc(x)-1) = csc^2(x) - 1 =
cot^2(x)`

So `cot(x)/(csc(x)+1)*(csc(x)-1)/(csc(x)-1) =
cot(x)(csc(x)-1)/(csc^2(x)-1) = cot(x)(csc(x)-1)/(cot^2(x)) =
(csc(x)-1)/cot(x)`

so our left hand side
becomes

`(csc(x)-1)/cot(x) + (csc(x)+1)/cot(x)` Now we can
add because they have the same denominator.  We
get:

`(csc(x)-1+csc(x)+1)/(cot(x))` simplifying the
top
`(2csc(x))/(cot(x))` To simplify I use the inverse function definitions
`csc(x)=1/sin(x)` and `cot(x) = cos(x)/sin(x)` to
get

`2*1/sin(x)*1/(cos(x)/sin(x)) =
2*1/sin(x)*sin(x)/cos(x) = 2/cos(x) = 2sec(x)`

Which is
what we wanted to prove

`cot(x)/(csc(x)+1) +
(csc(x)+1)/cot(x) = 2sec(x)`