Question

(csc(B)+cot (B))(csc(B)-cot(B))

Answer 1

You have the pattern of "difference of squares"

(csc(B)+cot (B))(csc(B)-cot(B))
= csc^2 B - cot^2 B

But csc^ B = 1 + cot^2B , one of the identies from
sin^2 B + cos^2 B= 1

so csc^2 B - cot^2 B
= 1 + cot^2 B - cot^2 B = 1