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

Expert Answers
beckden eNotes educator| Certified Educator

`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)`